diff --git a/curriculum/challenges/portuguese/01-responsive-web-design/applied-visual-design/set-the-font-size-for-multiple-heading-elements.md b/curriculum/challenges/portuguese/01-responsive-web-design/applied-visual-design/set-the-font-size-for-multiple-heading-elements.md
index b9b410c7dbb..d7dba594595 100644
--- a/curriculum/challenges/portuguese/01-responsive-web-design/applied-visual-design/set-the-font-size-for-multiple-heading-elements.md
+++ b/curriculum/challenges/portuguese/01-responsive-web-design/applied-visual-design/set-the-font-size-for-multiple-heading-elements.md
@@ -11,7 +11,9 @@ dashedName: set-the-font-size-for-multiple-heading-elements
 
 A propriedade `font-size` é usada para especificar o quão grande será o texto em um determinado elemento. Essa propriedade pode ser usada em múltiplos elementos para criar uma consistência visual dos textos na página. Nesse desafio, você vai definir os tamanhos das tipografias das tags `h1` até `h6` para balancear os tamanhos dos cabeçalhos.
 
-# --instructions-- <p>Na tag <code>style</code>, defina a propriedade <code>font-size</code> das tags:</p>
+# --instructions--
+
+  <p>Na tag <code>style</code>, defina a propriedade <code>font-size</code> das tags:</p>
 
   <ul>
     <li><code>h1</code> para 68px.</li>
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md
index 21be3de807a..fba38fd0081 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md
@@ -20,7 +20,9 @@ Como base, se nos fosse dado apenas o primeiro termo de sequência, seria mais s
 
 Assim, obtemos as seguintes OPs para a sequência cúbica:
 
-$$\begin{array}{ll} OP(1, n) = 1          & 1, {\color{red}1}, 1, 1, \ldots     \\\\ OP(2, n) = 7n−6       & 1, 8, {\color{red}{15}}, \ldots     \\\\ OP(3, n) = 6n^2−11n+6 & 1, 8, 27, {\color{red}{58}}, \ldots \\\\ OP(4, n) = n^3        & 1, 8, 27, 64, 125, \ldots \end{array}$$
+$$\begin{array}{ll}   OP(1, n) = 1          & 1, {\color{red}1}, 1, 1, \ldots     \\\\
+  OP(2, n) = 7n−6       & 1, 8, {\color{red}{15}}, \ldots     \\\\   OP(3, n) = 6n^2−11n+6 & 1, 8, 27, {\color{red}{58}}, \ldots \\\\
+  OP(4, n) = n^3        & 1, 8, 27, 64, 125, \ldots \end{array}$$
 
 Claramente não existem BOPs para k ≥ 4. Considerando a soma dos FITs gerados pelos BOPs (indicados em $\color{red}{red}$ acima), obtemos 1 + 15 + 58 = 74. Considere a seguinte função de geração de polinômios de décimo grau:
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md
index 14fad9ffe68..eb760912be8 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md
@@ -15,7 +15,10 @@ Vamos $S(A)$ representar a soma dos elementos no conjunto A, de tamanho n. Vamos
 
 Se $S(A)$ for minimizado por um determinado n, vamos chamar de um conjunto de soma especial ideal. Os primeiros cinco conjuntos de somas especiais ideais são fornecidos abaixo.
 
-$$\begin{align} & n = 1: \\{1\\} \\\\ & n = 2: \\{1, 2\\} \\\\ & n = 3: \\{2, 3, 4\\} \\\\ & n = 4: \\{3, 5, 6, 7\\} \\\\ & n = 5: \\{6, 9, 11, 12, 13\\} \\\\ \end{align}$$
+$$\begin{align}   & n = 1: \\{1\\} \\\\
+  & n = 2: \\{1, 2\\} \\\\   & n = 3: \\{2, 3, 4\\} \\\\
+  & n = 4: \\{3, 5, 6, 7\\} \\\\   & n = 5: \\{6, 9, 11, 12, 13\\} \\\\
+\end{align}$$
 
 Parece que, para um determinado conjunto ideal, $A = \\{a_1, a_2, \ldots, a_n\\}$, o próximo conjunto ideal é do formato $B = \\{b, a_1 + b, a_2 + b, \ldots, a_n + b\\}$, onde b é o elemento do "meio" na linha anterior.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md
index 17fe6cfb7e5..8627e2852b7 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md
@@ -14,7 +14,9 @@ $$\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$$
 
 Para `n` = 4, há exatamente três soluções distintas:
 
-$$\begin{align} & \frac{1}{5} + \frac{1}{20} = \frac{1}{4}\\\\ \\\\ & \frac{1}{6} + \frac{1}{12} = \frac{1}{4}\\\\ \\\\ & \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \end{align}$$
+$$\begin{align}   & \frac{1}{5} + \frac{1}{20} = \frac{1}{4}\\\\
+  \\\\   & \frac{1}{6} + \frac{1}{12} = \frac{1}{4}\\\\
+  \\\\ & \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \end{align}$$
 
 Qual é o menor valor de `n` para o qual o número de soluções distintas excede um mil?
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-109-darts.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-109-darts.md
index 9bf77ce54d9..a010546902a 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-109-darts.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-109-darts.md
@@ -20,7 +20,12 @@ Há muitas variações de regras, mas, no jogo mais popular, os jogadores começ
 
 Quando um jogador consegue terminar na pontuação atual, ela é chamada de "check-out". O check-out mais alto vale 170: T20 T20 D25 (dois 20s triplos e um duplo bull). Há exatamente onze maneiras distintas de marcar uma pontuação de 6:
 
-$$\begin{array} \text{D3} &    &    \\\\ D1        & D2 &    \\\\ S2        & D2 &    \\\\ D2        & D1 &    \\\\ S4        & D1 &    \\\\ S1        & S1 & D2 \\\\ S1        & T1 & D1 \\\\ S1        & S3 & D1 \\\\ D1        & D1 & D1 \\\\ D1        & S2 & D1 \\\\ S2        & S2 & D1 \end{array}$$
+$$\begin{array}   \text{D3} &    &    \\\\
+  D1        & D2 &    \\\\   S2        & D2 &    \\\\
+  D2        & D1 &    \\\\   S4        & D1 &    \\\\
+  S1        & S1 & D2 \\\\   S1        & T1 & D1 \\\\
+  S1        & S3 & D1 \\\\   D1        & D1 & D1 \\\\
+  D1        & S2 & D1 \\\\ S2        & S2 & D1 \end{array}$$
 
 Observe que D1 D2 é considerado diferente de D2 D1, pois terminam em duplas diferentes. No entanto, a combinação S1 T1 D1 é considerada a mesma que T1 S1 D1. Além disso, não devemos incluir erros ao considerar as combinações; por exemplo, D3 é o mesmo que 0 D3 e 0 0 D3. Incrivelmente, no total, existem 42336 maneiras distintas de fazer checkout. De quantas maneiras diferentes um jogador pode finalizar com uma pontuação inferior a 100?
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md
index 97f28fe83e9..f7efaf1e720 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md
@@ -14,11 +14,16 @@ $$n × n × \ldots × n = n^{15}$$
 
 Mas usando um método "binário" você pode calculá-lo em seis multiplicações:
 
-$$\begin{align} & n × n = n^2\\\\ & n^2 × n^2 = n^4\\\\ & n^4 × n^4 = n^8\\\\ & n^8 × n^4 = n^{12}\\\\ & n^{12} × n^2 = n^{14}\\\\ & n^{14} × n = n^{15} \end{align}$$
+$$\begin{align}   & n × n = n^2\\\\
+  & n^2 × n^2 = n^4\\\\   & n^4 × n^4 = n^8\\\\
+  & n^8 × n^4 = n^{12}\\\\   & n^{12} × n^2 = n^{14}\\\\
+  & n^{14} × n = n^{15} \end{align}$$
 
 No entanto, ainda é possível calculá-lo em apenas cinco multiplicações:
 
-$$\begin{align} & n × n = n^2\\\\ & n^2 × n = n^3\\\\ & n^3 × n^3 = n^6\\\\ & n^6 × n^6 = n^{12}\\\\ & n^{12} × n^3 = n^{15} \end{align}$$
+$$\begin{align}   & n × n = n^2\\\\
+  & n^2 × n = n^3\\\\   & n^3 × n^3 = n^6\\\\
+  & n^6 × n^6 = n^{12}\\\\ & n^{12} × n^3 = n^{15} \end{align}$$
 
 Definiremos $m(k)$ como o número mínimo de multiplicações para calcular $n^k$; por exemplo, $m(15) = 5$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md
index cf0f9558d5d..053f390865d 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md
@@ -14,7 +14,8 @@ Para este problema, estaremos interessados em valores de $x$ para os quais $A_{F
 
 Surpreendentemente,
 
-$$\begin{align} A_F(\frac{1}{2}) & = (\frac{1}{2}) × 1 + {(\frac{1}{2})}^2 × 1 + {(\frac{1}{2})}^3 × 2 + {(\frac{1}{2})}^4 × 3 + {(\frac{1}{2})}^5 × 5 + \cdots \\\\ & = \frac{1}{2} + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{5}{32} + \cdots \\\\ & = 2 \end{align}$$
+$$\begin{align} A_F(\frac{1}{2}) & = (\frac{1}{2}) × 1 + {(\frac{1}{2})}^2 × 1 + {(\frac{1}{2})}^3 × 2 + {(\frac{1}{2})}^4 × 3 + {(\frac{1}{2})}^5 × 5 + \cdots \\\\
+                 & = \frac{1}{2} + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{5}{32} + \cdots \\\\ & = 2 \end{align}$$
 
 Os valores correspondentes de $x$ para os primeiros cinco números naturais são mostrados abaixo.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md
index f4fa41376bd..54c2905cc6a 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md
@@ -17,13 +17,17 @@ No exemplo abaixo, pode ser facilmente verificado que o triângulo marcado satis
 
 Queremos fazer uma matriz triangular desse tipo com mil fileiras. Então, geramos 500500 números pseudoaleatórios $s_k$ no intervalo $±2^{19}$, usando um tipo de gerador de número aleatório (conhecido como gerador congruente linear), da seguinte forma:
 
-$$\begin{align} t := & \\ 0\\\\ \text{for}\\ & k = 1\\ \text{up to}\\ k = 500500:\\\\ & t := (615949 × t + 797807)\\ \text{modulo}\\ 2^{20}\\\\ & s_k := t − 219\\\\ \end{align}$$
+$$\begin{align}   t := & \\ 0\\\\
+  \text{for}\\ & k = 1\\ \text{up to}\\ k = 500500:\\\\   & t := (615949 × t + 797807)\\ \text{modulo}\\ 2^{20}\\\\
+  & s_k := t − 219\\\\ \end{align}$$
 
 Assim: $s_1 = 273519$, $s_2 = −153582$, $s_3 = 450905$ e assim por diante.
 
 Nossa matriz triangular é então formada usando os pseudonúmeros aleatórios, ou seja:
 
-$$ s_1 \\\\ s_2\\;s_3 \\\\ s_4\\; s_5\\; s_6 \\\\ s_7\\; s_8\\; s_9\\; s_{10} \\\\ \ldots $$
+$$ s_1 \\\\
+s_2\\;s_3 \\\\ s_4\\; s_5\\; s_6 \\\\
+s_7\\; s_8\\; s_9\\; s_{10} \\\\ \ldots $$
 
 Os subtriângulos podem começar em qualquer elemento da matriz e se estender até onde quisermos (pegando os dois elementos diretamente abaixo dele na próxima fileira, sendo os três elementos diretamente abaixo da linha depois disso e assim por diante).
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md
index a80823652c8..511ddfa54c6 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md
@@ -12,7 +12,10 @@ Considere a equação diofantina $\frac{1}{a} + \frac{1}{b} = \frac{p}{{10}^n}$,
 
 Para $n = 1$, esta equação tem 20 soluções listadas abaixo:
 
-$$\begin{array}{lllll} \frac{1}{1}  + \frac{1}{1}  = \frac{20}{10} & \frac{1}{1} + \frac{1}{2}  = \frac{15}{10} & \frac{1}{1}  + \frac{1}{5}  = \frac{12}{10} & \frac{1}{1} + \frac{1}{10} = \frac{11}{10} & \frac{1}{2}  + \frac{1}{2}  = \frac{10}{10} \\\\ \frac{1}{2}  + \frac{1}{5}  = \frac{7}{10}   & \frac{1}{2} + \frac{1}{10} = \frac{6}{10} & \frac{1}{3}  + \frac{1}{6}  = \frac{5}{10}   & \frac{1}{3} + \frac{1}{15} = \frac{4}{10} & \frac{1}{4}  + \frac{1}{4}  = \frac{5}{10} \\\\ \frac{1}{4}  + \frac{1}{4}  = \frac{5}{10}  & \frac{1}{5}  + \frac{1}{5}  = \frac{4}{10} & \frac{1}{5}  + \frac{1}{10} = \frac{3}{10}  & \frac{1}{6}  + \frac{1}{30} = \frac{2}{10} & \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \\\\ \frac{1}{11} + \frac{1}{110} = \frac{1}{10} & \frac{1}{12} + \frac{1}{60}  = \frac{1}{10} & \frac{1}{14} + \frac{1}{35}  = \frac{1}{10} & \frac{1}{15} + \frac{1}{30}  = \frac{1}{10} & \frac{1}{20} + \frac{1}{20}  = \frac{1}{10} \end{array}$$
+$$\begin{array}{lllll} \frac{1}{1}  + \frac{1}{1}  = \frac{20}{10} & \frac{1}{1} + \frac{1}{2}  = \frac{15}{10} & \frac{1}{1}  + \frac{1}{5}  = \frac{12}{10} & \frac{1}{1} + \frac{1}{10} = \frac{11}{10} & \frac{1}{2}  + \frac{1}{2}  = \frac{10}{10} \\\\
+  \frac{1}{2}  + \frac{1}{5}  = \frac{7}{10}   & \frac{1}{2} + \frac{1}{10} = \frac{6}{10} & \frac{1}{3}  + \frac{1}{6}  = \frac{5}{10}   & \frac{1}{3} + \frac{1}{15} = \frac{4}{10} & \frac{1}{4}  + \frac{1}{4}  = \frac{5}{10} \\\\
+  \frac{1}{4}  + \frac{1}{4}  = \frac{5}{10}  & \frac{1}{5}  + \frac{1}{5}  = \frac{4}{10} & \frac{1}{5}  + \frac{1}{10} = \frac{3}{10}  & \frac{1}{6}  + \frac{1}{30} = \frac{2}{10} & \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \\\\
+  \frac{1}{11} + \frac{1}{110} = \frac{1}{10} & \frac{1}{12} + \frac{1}{60}  = \frac{1}{10} & \frac{1}{14} + \frac{1}{35}  = \frac{1}{10} & \frac{1}{15} + \frac{1}{30}  = \frac{1}{10} & \frac{1}{20} + \frac{1}{20}  = \frac{1}{10} \end{array}$$
 
 Quantas soluções tem esta equação para $1 ≤ n ≤ 9$?
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md
index 11c93a6a7d5..c16e5997dbf 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md
@@ -12,7 +12,10 @@ Um número composto pode ser fatorado de várias maneiras.
 
 Por exemplo, não incluindo a multiplicação por um, 24 podem ser fatorado de 7 formas distintas:
 
-$$\begin{align} & 24 = 2 \times 2 \times 2 \times 3\\\\ & 24 = 2 \times 3 \times 4  \\\\ & 24 = 2 \times 2 \times 6  \\\\ & 24 = 4 \times 6    \\\\ & 24 = 3 \times 8    \\\\ & 24 = 2 \times 12   \\\\ & 24 = 24 \end{align}$$
+$$\begin{align}   & 24 = 2 \times 2 \times 2 \times 3\\\\
+  & 24 = 2 \times 3 \times 4  \\\\   & 24 = 2 \times 2 \times 6  \\\\
+  & 24 = 4 \times 6    \\\\   & 24 = 3 \times 8    \\\\
+  & 24 = 2 \times 12   \\\\ & 24 = 24 \end{align}$$
 
 Lembre-se de que a raiz de algarismos de um número, na base 10, é encontrada adicionando os algarismos daquele número e repetindo esse processo até que um número chegue a menos de 10. Assim, a raiz dos algarismos de 467 é 8.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md
index d5d4094d0fd..48ac9f7b52c 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md
@@ -12,7 +12,8 @@ Para qualquer $N$, considere $f(N)$ como os últimos cinco algarismos antes dos
 
 Por exemplo:
 
-$$\begin{align} & 9! = 362880 \\; \text{so} \\; f(9) = 36288 \\\\ & 10! = 3628800 \\; \text{so} \\; f(10) = 36288 \\\\ & 20! = 2432902008176640000 \\; \text{so} \\; f(20) = 17664 \end{align}$$
+$$\begin{align}   & 9! = 362880 \\; \text{so} \\; f(9) = 36288 \\\\
+  & 10! = 3628800 \\; \text{so} \\; f(10) = 36288 \\\\ & 20! = 2432902008176640000 \\; \text{so} \\; f(20) = 17664 \end{align}$$
 
 Encontre $f(1.000.000.000.000)$
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md
index 610b3f699c3..e05e5ed3b0a 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md
@@ -18,7 +18,8 @@ Agora, nesse triângulo, podem ser observados dezesseis triângulos de forma, ta
 
 Se quisermos indicar que $T(n)$ é o número de triângulos presentes em um triângulo de tamanho $n$, então
 
-$$\begin{align} & T(1) = 16 \\\\ & T(2) = 104 \end{align}$$
+$$\begin{align}   & T(1) = 16 \\\\
+  & T(2) = 104 \end{align}$$
 
 Encontre $T(36)$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-165-intersections.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-165-intersections.md
index 4fea2e94b83..5ed5ba9f06d 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-165-intersections.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-165-intersections.md
@@ -16,13 +16,17 @@ Chamaremos de ponto comum $T$ de dois segmentos $L_1$ e $L_2$ um ponto de inters
 
 Considere os três segmentos $L_1$, $L_2$, e $L_3$:
 
-$$\begin{align} & L_1: (27, 44) \\;\text{to}\\; (12, 32) \\\\ & L_2: (46, 53) \\;\text{to}\\; (17, 62) \\\\ & L_3: (46, 70) \\;\text{to}\\; (22, 40) \\\\ \end{align}$$
+$$\begin{align}   & L_1: (27, 44) \\;\text{to}\\; (12, 32) \\\\
+  & L_2: (46, 53) \\;\text{to}\\; (17, 62) \\\\   & L_3: (46, 70) \\;\text{to}\\; (22, 40) \\\\
+\end{align}$$
 
 É possível verificar que os segmentos de reta $L_2$ e $L_3$ têm um ponto de interseção verdadeira. Percebemos que, como um dos pontos de extremidade de $L_3$: (22, 40) fica sobre $L_1$, este não é considerado um ponto de interseção verdadeira. $L_1$ e $L_2$ não têm um ponto em comum. Portanto, entre os três segmentos de reta, encontramos um ponto de interseção verdadeira.
 
 Façamos agora o mesmo em 5.000 segmentos de reta. Para isso, geramos 20.000 números usando o chamado gerador pseudoaleatório de números "Blum Blum Shub".
 
-$$\begin{align} & s_0 = 290797 \\\\ & s_{n + 1} = s_n × s_n (\text{modulo}\\; 50515093) \\\\ & t_n = s_n (\text{modulo}\\; 500) \\\\ \end{align}$$
+$$\begin{align}   & s_0 = 290797 \\\\
+  & s_{n + 1} = s_n × s_n (\text{modulo}\\; 50515093) \\\\   & t_n = s_n (\text{modulo}\\; 500) \\\\
+\end{align}$$
 
 Para criar cada segmento de reta, usamos quatro números consecutivos $t_n$. Ou seja, o primeiro segmento de reta é dado por:
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md
index 04b03ef312e..04e11d6e03d 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md
@@ -12,7 +12,9 @@ Uma grade de 4x4 é preenchida por algarismos $d$, sendo que $0 ≤ d ≤ 9$.
 
 Pode-se ver que na grade
 
-$$\begin{array}{} 6 & 3 & 3 & 0 \\\\ 5 & 0 & 4 & 3 \\\\ 0 & 7 & 1 & 4 \\\\ 1 & 2 & 4 & 5 \end{array}$$
+$$\begin{array}{}   6 & 3 & 3 & 0 \\\\
+  5 & 0 & 4 & 3 \\\\   0 & 7 & 1 & 4 \\\\
+  1 & 2 & 4 & 5 \end{array}$$
 
 a soma de cada linha e de cada coluna tem o valor 12. Além disso, a soma de cada diagonal também é 12.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
index d07c0c96b38..318e503ab59 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
@@ -14,7 +14,9 @@ Defina $f(0)=1$ e $f(n)$ como o número de diferentes maneiras pelas quais $n$ p
 
 Por exemplo, $f(10)=5$ já que há cinco maneiras diferentes de expressar 10:
 
-$$\begin{align} & 1 + 1 + 8 \\\\ & 1 + 1 + 4 + 4 \\\\ & 1 + 1 + 2 + 2 + 4 \\\\ & 2 + 4 + 4 \\\\ & 2 + 8 \end{align}$$
+$$\begin{align}   & 1 + 1 + 8 \\\\
+  & 1 + 1 + 4 + 4 \\\\   & 1 + 1 + 2 + 2 + 4 \\\\
+  & 2 + 4 + 4 \\\\ & 2 + 8 \end{align}$$
 
 Qual é $f({10}^{25})$?
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md
index fc127d6f2a1..1fcb7cf9a27 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md
@@ -12,7 +12,8 @@ dashedName: >-
 
 Pegue o número 6 e multiplique-o por 1273 e 9854:
 
-$$\begin{align} & 6 × 1273 = 7638 \\\\ & 6 × 9854 = 59124 \\\\ \end{align}$$
+$$\begin{align}   & 6 × 1273 = 7638 \\\\
+  & 6 × 9854 = 59124 \\\\ \end{align}$$
 
 Ao concatenar esses produtos, temos o pandigital de 1 a 9 763859124. Chamaremos 763859124 de "produto concatenado de 6 e (1273, 9854)". Observe, também, que a concatenação dos números de entrada, 612739854, também é um pandigital de 1 a 9.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md
index 988a1a80eab..1c7036458f8 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md
@@ -12,7 +12,9 @@ dashedName: >-
 
 Para um número inteiro positivo $n$, considere $f(n)$ como a soma dos quadrados dos algarismos (na base 10) de $n$, por exemplo,
 
-$$\begin{align} & f(3) = 3^2 = 9 \\\\ & f(25) = 2^2 + 5^2 = 4 + 25 = 29 \\\\ & f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36 \\\\ \end{align}$$
+$$\begin{align}   & f(3) = 3^2 = 9 \\\\
+  & f(25) = 2^2 + 5^2 = 4 + 25 = 29 \\\\   & f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36 \\\\
+\end{align}$$
 
 Encontre os últimos nove algarismos da soma de todos os $n$, sendo que $0 &lt; n &lt; {10}^{20}$, de modo que $f(n)$ seja um quadrado perfeito.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md
index 358e4cfe347..d880d0cf5d2 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md
@@ -10,7 +10,8 @@ dashedName: problem-180-rational-zeros-of-a-function-of-three-variables
 
 Para qualquer número inteiro $n$, considere as três funções
 
-$$\begin{align} & f_{1,n}(x,y,z) = x^{n + 1} + y^{n + 1} − z^{n + 1}\\\\ & f_{2,n}(x,y,z) = (xy + yz + zx) \times (x^{n - 1} + y^{n - 1} − z^{n - 1})\\\\ & f_{3,n}(x,y,z) = xyz \times (x^{n - 2} + y^{n - 2} − z^{n - 2}) \end{align}$$
+$$\begin{align}   & f_{1,n}(x,y,z) = x^{n + 1} + y^{n + 1} − z^{n + 1}\\\\
+  & f_{2,n}(x,y,z) = (xy + yz + zx) \times (x^{n - 1} + y^{n - 1} − z^{n - 1})\\\\ & f_{3,n}(x,y,z) = xyz \times (x^{n - 2} + y^{n - 2} − z^{n - 2}) \end{align}$$
 
 e suas combinações
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-185-number-mind.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-185-number-mind.md
index 78e3b3a0ce9..523c456a548 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-185-number-mind.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-185-number-mind.md
@@ -14,13 +14,27 @@ Em vez de peças coloridas, você tem que adivinhar uma sequência secreta de al
 
 Por exemplo, dados os seguintes palpites para uma sequência secreta de 5 algarismos
 
-$$\begin{align} & 90342 ;2\\;\text{correct}\\\\ & 70794 ;0\\;\text{correct}\\\\ & 39458 ;2\\;\text{correct}\\\\ & 34109 ;1\\;\text{correct}\\\\ & 51545 ;2\\;\text{correct}\\\\ & 12531 ;1\\;\text{correct} \end{align}$$
+$$\begin{align}   & 90342 ;2\\;\text{correct}\\\\
+  & 70794 ;0\\;\text{correct}\\\\   & 39458 ;2\\;\text{correct}\\\\
+  & 34109 ;1\\;\text{correct}\\\\   & 51545 ;2\\;\text{correct}\\\\
+  & 12531 ;1\\;\text{correct} \end{align}$$
 
 A sequência correta 39542 é única.
 
 Com base nos palpites abaixo
 
-$$\begin{align} & 5616185650518293 ;2\\;\text{correct}\\\\ & 3847439647293047 ;1\\;\text{correct}\\\\ & 5855462940810587 ;3\\;\text{correct}\\\\ & 9742855507068353 ;3\\;\text{correct}\\\\ & 4296849643607543 ;3\\;\text{correct}\\\\ & 3174248439465858 ;1\\;\text{correct}\\\\ & 4513559094146117 ;2\\;\text{correct}\\\\ & 7890971548908067 ;3\\;\text{correct}\\\\ & 8157356344118483 ;1\\;\text{correct}\\\\ & 2615250744386899 ;2\\;\text{correct}\\\\ & 8690095851526254 ;3\\;\text{correct}\\\\ & 6375711915077050 ;1\\;\text{correct}\\\\ & 6913859173121360 ;1\\;\text{correct}\\\\ & 6442889055042768 ;2\\;\text{correct}\\\\ & 2321386104303845 ;0\\;\text{correct}\\\\ & 2326509471271448 ;2\\;\text{correct}\\\\ & 5251583379644322 ;2\\;\text{correct}\\\\ & 1748270476758276 ;3\\;\text{correct}\\\\ & 4895722652190306 ;1\\;\text{correct}\\\\ & 3041631117224635 ;3\\;\text{correct}\\\\ & 1841236454324589 ;3\\;\text{correct}\\\\ & 2659862637316867 ;2\\;\text{correct} \end{align}$$
+$$\begin{align}   & 5616185650518293 ;2\\;\text{correct}\\\\
+  & 3847439647293047 ;1\\;\text{correct}\\\\   & 5855462940810587 ;3\\;\text{correct}\\\\
+  & 9742855507068353 ;3\\;\text{correct}\\\\   & 4296849643607543 ;3\\;\text{correct}\\\\
+  & 3174248439465858 ;1\\;\text{correct}\\\\   & 4513559094146117 ;2\\;\text{correct}\\\\
+  & 7890971548908067 ;3\\;\text{correct}\\\\   & 8157356344118483 ;1\\;\text{correct}\\\\
+  & 2615250744386899 ;2\\;\text{correct}\\\\   & 8690095851526254 ;3\\;\text{correct}\\\\
+  & 6375711915077050 ;1\\;\text{correct}\\\\   & 6913859173121360 ;1\\;\text{correct}\\\\
+  & 6442889055042768 ;2\\;\text{correct}\\\\   & 2321386104303845 ;0\\;\text{correct}\\\\
+  & 2326509471271448 ;2\\;\text{correct}\\\\   & 5251583379644322 ;2\\;\text{correct}\\\\
+  & 1748270476758276 ;3\\;\text{correct}\\\\   & 4895722652190306 ;1\\;\text{correct}\\\\
+  & 3041631117224635 ;3\\;\text{correct}\\\\   & 1841236454324589 ;3\\;\text{correct}\\\\
+  & 2659862637316867 ;2\\;\text{correct} \end{align}$$
 
 Encontre a sequência secreta única de 16 algarismos.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md
index a885946bb28..5e987413281 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md
@@ -10,7 +10,13 @@ dashedName: problem-196-prime-triplets
 
 Construa um triângulo com todos os números inteiros positivos da seguinte maneira:
 
-$$\begin{array}{rrr} &  1 \\\\ &  \color{red}{2} &  \color{red}{3} \\\\ &  4 & \color{red}{5} &  6 \\\\ &  \color{red}{7} &  8 &  9 & 10 \\\\ & \color{red}{11} & 12 & \color{red}{13} & 14 & 15  \\\\ & 16 & \color{red}{17} & 18 & \color{red}{19} & 20 & 21 \\\\ & 22 & \color{red}{23} & 24 & 25 & 26 & 27 & 28 \\\\ & \color{red}{29} & 30 & \color{red}{31} & 32 & 33 & 34 & 35 & 36 \\\\ & \color{red}{37} & 38 & 39 & 40 & \color{red}{41} & 42 & \color{red}{43} & 44 & 45 \\\\ & 46 & \color{red}{47} & 48 & 49 & 50 & 51 & 52 & \color{red}{53} & 54 & 55 \\\\ & 56 & 57 & 58 & \color{red}{59} & 60 & \color{red}{61} & 62 & 63 & 64 & 65 & 66 \\\\ & \cdots \end{array}$$
+$$\begin{array}{rrr}   &  1 \\\\
+  &  \color{red}{2} &  \color{red}{3} \\\\   &  4 & \color{red}{5} &  6 \\\\
+  &  \color{red}{7} &  8 &  9 & 10 \\\\   & \color{red}{11} & 12 & \color{red}{13} & 14 & 15  \\\\
+  & 16 & \color{red}{17} & 18 & \color{red}{19} & 20 & 21 \\\\   & 22 & \color{red}{23} & 24 & 25 & 26 & 27 & 28 \\\\
+  & \color{red}{29} & 30 & \color{red}{31} & 32 & 33 & 34 & 35 & 36 \\\\   & \color{red}{37} & 38 & 39 & 40 & \color{red}{41} & 42 & \color{red}{43} & 44 & 45 \\\\
+  & 46 & \color{red}{47} & 48 & 49 & 50 & 51 & 52 & \color{red}{53} & 54 & 55 \\\\   & 56 & 57 & 58 & \color{red}{59} & 60 & \color{red}{61} & 62 & 63 & 64 & 65 & 66 \\\\
+  & \cdots \end{array}$$
 
 Cada número inteiro positivo tem até oito vizinhos no triângulo.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md
index c54e71586dc..8a53e0b39f3 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md
@@ -12,7 +12,17 @@ Para qualquer conjunto $A$ de números, considere $sum(A)$ a soma dos elementos
 
 Considere o conjunto $B = \\{1,3,6,8,10,11\\}$. Há 20 subconjuntos de $B$ contendo três elementos, e suas somas são:
 
-$$\begin{align} & sum(\\{1,3,6\\}) = 10 \\\\ & sum(\\{1,3,8\\}) = 12 \\\\ & sum(\\{1,3,10\\}) = 14 \\\\ & sum(\\{1,3,11\\}) = 15 \\\\ & sum(\\{1,6,8\\}) = 15 \\\\ & sum(\\{1,6,10\\}) = 17 \\\\ & sum(\\{1,6,11\\}) = 18 \\\\ & sum(\\{1,8,10\\}) = 19 \\\\ & sum(\\{1,8,11\\}) = 20 \\\\ & sum(\\{1,10,11\\}) = 22 \\\\ & sum(\\{3,6,8\\}) = 17 \\\\ & sum(\\{3,6,10\\}) = 19 \\\\ & sum(\\{3,6,11\\}) = 20 \\\\ & sum(\\{3,8,10\\}) = 21 \\\\ & sum(\\{3,8,11\\}) = 22 \\\\ & sum(\\{3,10,11\\}) = 24 \\\\ & sum(\\{6,8,10\\}) = 24 \\\\ & sum(\\{6,8,11\\}) = 25 \\\\ & sum(\\{6,10,11\\}) = 27 \\\\ & sum(\\{8,10,11\\}) = 29 \\end{align}$$
+$$\begin{align}   & sum(\\{1,3,6\\}) = 10 \\\\
+  & sum(\\{1,3,8\\}) = 12 \\\\   & sum(\\{1,3,10\\}) = 14 \\\\
+  & sum(\\{1,3,11\\}) = 15 \\\\   & sum(\\{1,6,8\\}) = 15 \\\\
+  & sum(\\{1,6,10\\}) = 17 \\\\   & sum(\\{1,6,11\\}) = 18 \\\\
+  & sum(\\{1,8,10\\}) = 19 \\\\   & sum(\\{1,8,11\\}) = 20 \\\\
+  & sum(\\{1,10,11\\}) = 22 \\\\   & sum(\\{3,6,8\\}) = 17 \\\\
+  & sum(\\{3,6,10\\}) = 19 \\\\   & sum(\\{3,6,11\\}) = 20 \\\\
+  & sum(\\{3,8,10\\}) = 21 \\\\   & sum(\\{3,8,11\\}) = 22 \\\\
+  & sum(\\{3,10,11\\}) = 24 \\\\   & sum(\\{6,8,10\\}) = 24 \\\\
+  & sum(\\{6,8,11\\}) = 25 \\\\   & sum(\\{6,10,11\\}) = 27 \\\\
+  & sum(\\{8,10,11\\}) = 29 \\end{align}$$
 
 Algumas destas somas ocorrem mais de uma vez, outras são únicas. Para um conjunto de $A$, considere $U(A,k)$ como sendo o conjunto de somas únicas de subconjuntos de $k$ elementos de $A$, No nosso exemplo, encontramos $U(B,3) = \\{10,12,14,18,21,25,27,29\\}$ e $sum(U(B,3)) = 156$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md
index 1de46472826..69e16acb744 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md
@@ -10,7 +10,11 @@ dashedName: problem-203-squarefree-binomial-coefficients
 
 Os coeficientes binomiais $\displaystyle\binom{n}{k}$ podem ser organizados em forma triangular, no triângulo de Pascal, assim:
 
-$$\begin{array}{ccccccccccccccc} &   &   &   &    &    &    &  1 &    &    &    &   &   &   &   \\\\ &   &   &   &    &    &  1 &    & 1  &    &    &   &   &   &   \\\\ &   &   &   &    &  1 &    &  2 &    &  1 &    &   &   &   &   \\\\ &   &   &   &  1 &    &  3 &    &  3 &    &  1 &   &   &   &   \\\\ &   &   & 1 &    &  4 &    &  6 &    &  4 &    & 1 &   &   &   \\\\ &   & 1 &   &  5 &    & 10 &    & 10 &    &  5 &   & 1 &   &   \\\\ & 1 &   & 6 &    & 15 &    & 20 &    & 15 &    & 6 &   & 1 &   \\\\ 1 &   & 7 &   & 21 &    & 35 &    & 35 &    & 21 &   & 7 &   & 1 \\\\ &   &   &   &    &    &    & \ldots \end{array}$$
+$$\begin{array}{ccccccccccccccc}    &   &   &   &    &    &    &  1 &    &    &    &   &   &   &   \\\\
+   &   &   &   &    &    &  1 &    & 1  &    &    &   &   &   &   \\\\    &   &   &   &    &  1 &    &  2 &    &  1 &    &   &   &   &   \\\\
+   &   &   &   &  1 &    &  3 &    &  3 &    &  1 &   &   &   &   \\\\    &   &   & 1 &    &  4 &    &  6 &    &  4 &    & 1 &   &   &   \\\\
+   &   & 1 &   &  5 &    & 10 &    & 10 &    &  5 &   & 1 &   &   \\\\    & 1 &   & 6 &    & 15 &    & 20 &    & 15 &    & 6 &   & 1 &   \\\\
+ 1 &   & 7 &   & 21 &    & 35 &    & 35 &    & 21 &   & 7 &   & 1 \\\\ &   &   &   &    &    &    & \ldots \end{array}$$
 
 Podemos ver que as primeiras oito linhas do triângulo de Pascal contêm doze números distintos: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 e 35.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md
index 440218de4f1..580268c0b3f 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md
@@ -20,7 +20,11 @@ Assim, $P(6) = \frac{1}{2}$.
 
 Na tabela a seguir estão listados alguns valores de $P(m)$
 
-$$\begin{align} & P(5) = \frac{1}{1}    \\\\ & P(10) = \frac{1}{2}   \\\\ & P(15) = \frac{2}{3}   \\\\ & P(20) = \frac{1}{2}   \\\\ & P(25) = \frac{1}{2}   \\\\ & P(30) = \frac{2}{5}   \\\\ & \ldots                \\\\ & P(180) = \frac{1}{4}  \\\\ & P(185) = \frac{3}{13} \end{align}$$
+$$\begin{align}   & P(5) = \frac{1}{1}    \\\\
+  & P(10) = \frac{1}{2}   \\\\   & P(15) = \frac{2}{3}   \\\\
+  & P(20) = \frac{1}{2}   \\\\   & P(25) = \frac{1}{2}   \\\\
+  & P(30) = \frac{2}{5}   \\\\   & \ldots                \\\\
+  & P(180) = \frac{1}{4}  \\\\ & P(185) = \frac{3}{13} \end{align}$$
 
 Encontre o menor $m$ para o qual $P(m) &lt; \frac{1}{12.345}$
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md
index 0a48b57aaf3..d7820202a33 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md
@@ -12,7 +12,10 @@ Um cuboide alinhado em seus eixos, especificado pelos parâmetros $\{ (x_0,y_0,z
 
 Considere $C_1, \ldots, C_{50000}$ como sendo uma coleção de 50.000 cuboides alinhados em seus eixos, de modo que $C_n$ tenha parâmetros
 
-$$\begin{align} & x_0 = S_{6n - 5} \\; \text{modulo} \\; 10000    \\\\ & y_0 = S_{6n - 4} \\; \text{modulo} \\; 10000    \\\\ & z_0 = S_{6n - 3} \\; \text{modulo} \\; 10000    \\\\ & dx = 1 + (S_{6n - 2} \\; \text{modulo} \\; 399) \\\\ & dy = 1 + (S_{6n - 1} \\; \text{modulo} \\; 399) \\\\ & dz = 1 + (S_{6n} \\; \text{modulo} \\; 399)     \\\\ \end{align}$$
+$$\begin{align}   & x_0 = S_{6n - 5} \\; \text{modulo} \\; 10000    \\\\
+  & y_0 = S_{6n - 4} \\; \text{modulo} \\; 10000    \\\\   & z_0 = S_{6n - 3} \\; \text{modulo} \\; 10000    \\\\
+  & dx = 1 + (S_{6n - 2} \\; \text{modulo} \\; 399) \\\\   & dy = 1 + (S_{6n - 1} \\; \text{modulo} \\; 399) \\\\
+  & dz = 1 + (S_{6n} \\; \text{modulo} \\; 399)     \\\\ \end{align}$$
 
 onde $S_1, \ldots, S_{300000}$ vem do "Gerador Fibonacci com atraso":
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md
index 9675f70327d..cb613fd937c 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md
@@ -12,7 +12,11 @@ Considere $φ$ como sendo a função totiente de Euler, ou seja, para um número
 
 Ao iterar por $φ$, cada número inteiro positivo gera uma cadeia decrescente de números terminando em 1. Ex: se começarmos com 5 a sequência 5,4,2,1 é gerada. Aqui está uma lista de todas as cadeias com comprimento 4:
 
-$$\begin{align} 5,4,2,1 & \\\\ 7,6,2,1 & \\\\ 8,4,2,1 & \\\\ 9,6,2,1 & \\\\ 10,4,2,1 & \\\\ 12,4,2,1 & \\\\ 14,6,2,1 & \\\\ 18,6,2,1 & \end{align}$$
+$$\begin{align}    5,4,2,1 & \\\\
+   7,6,2,1 & \\\\    8,4,2,1 & \\\\
+   9,6,2,1 & \\\\   10,4,2,1 & \\\\
+  12,4,2,1 & \\\\   14,6,2,1 & \\\\
+  18,6,2,1 & \end{align}$$
 
 Apenas duas dessas cadeias começam com um número primo e sua soma é 12.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md
index 324c0d16739..fe2da5bfa14 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md
@@ -10,7 +10,8 @@ dashedName: problem-228-minkowski-sums
 
 Considere $S_n$ como o polígono – ou forma – regular de $n$ lados, cujos vértices $v_k (k = 1, 2, \ldots, n)$ têm as coordenadas:
 
-$$\begin{align} & x_k = cos(\frac{2k - 1}{n} × 180°) \\\\ & y_k = sin(\frac{2k - 1}{n} × 180°) \end{align}$$
+$$\begin{align}   & x_k = cos(\frac{2k - 1}{n} × 180°) \\\\
+  & y_k = sin(\frac{2k - 1}{n} × 180°) \end{align}$$
 
 Cada $S_n$ deve ser interpretado como uma forma preenchida que consiste em todos os pontos no perímetro e no interior.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md
index c0dcee7b6d0..1700b470689 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md
@@ -10,13 +10,17 @@ dashedName: problem-229-four-representations-using-squares
 
 Considere o número 3600. Ele é muito especial, porque
 
-$$\begin{align} & 3600 = {48}^2 + {36}^2   \\\\ & 3600 = {20}^2 + {2×40}^2 \\\\ & 3600 = {30}^2 + {3×30}^2 \\\\ & 3600 = {45}^2 + {7×15}^2 \\\\ \end{align}$$
+$$\begin{align}   & 3600 = {48}^2 + {36}^2   \\\\
+  & 3600 = {20}^2 + {2×40}^2 \\\\   & 3600 = {30}^2 + {3×30}^2 \\\\
+  & 3600 = {45}^2 + {7×15}^2 \\\\ \end{align}$$
 
 Da mesma forma, descobrimos que $88201 = {99}^2 + {280}^2 = {287}^2 + 2 × {54}^2 = {283}^2 + 3 × {52}^2 = {197}^2 + 7 × {84}^2$.
 
 Em 1747, Euler provou quais números são representáveis como uma soma de dois quadrados. Estamos interessados nos números $n$ que admitem representações de todos os quatro tipos a seguir:
 
-$$\begin{align} & n = {a_1}^2 + {b_1}^2  \\\\ & n = {a_2}^2 + 2{b_2}^2 \\\\ & n = {a_3}^2 + 3{b_3}^2 \\\\ & n = {a_7}^2 + 7{b_7}^2 \\\\ \end{align}$$
+$$\begin{align}   & n = {a_1}^2 + {b_1}^2  \\\\
+  & n = {a_2}^2 + 2{b_2}^2 \\\\   & n = {a_3}^2 + 3{b_3}^2 \\\\
+  & n = {a_7}^2 + 7{b_7}^2 \\\\ \end{align}$$
 
 onde $a_k$ e $b_k$ são números inteiros positivos.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md
index c529532c4b3..3f97c474b2f 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md
@@ -18,17 +18,21 @@ Considere $A = 1.415.926.535$, $B = 8.979.323.846$. Queremos encontrar, digamos,
 
 Os primeiros termos de $F_{A,B}$ são:
 
-$$\begin{align} & 1.415.926\\,535 \\\\ & 8.979.323.846 \\\\ & 14.159.265.358.979.323.846 \\\\ & 897.932.384.614.159.265.358.979.323.846 \\\\ & 14.159.265.358.979.323.846.897.932.384.614.15\color{red}{9}.265.358.979.323.846 \end{align}$$
+$$\begin{align}   & 1.415.926\\,535 \\\\
+  & 8.979.323.846 \\\\   & 14.159.265.358.979.323.846 \\\\
+  & 897.932.384.614.159.265.358.979.323.846 \\\\ & 14.159.265.358.979.323.846.897.932.384.614.15\color{red}{9}.265.358.979.323.846 \end{align}$$
 
 Então, $D_{A,B}(35)$ é o ${35}^{\text{o}}$ algarismo no quinto termo, que é 9.
 
 Agora, usamos para $A$ os primeiros 100 algarismos de $π$ antes do ponto decimal:
 
-$$\begin{align} & 14.159.265.358.979.323.846.264.338.327.950.288.419.716.939.937.510 \\\\ & 58.209.749.445.923.078.164.062.862.089.986.280.348.253.421.170.679 \end{align}$$
+$$\begin{align}   & 14.159.265.358.979.323.846.264.338.327.950.288.419.716.939.937.510 \\\\
+  & 58.209.749.445.923.078.164.062.862.089.986.280.348.253.421.170.679 \end{align}$$
 
 e para $B$ os próximos cem algarismos:
 
-$$\begin{align} & 82.148.086.513.282.306.647.093.844.609.550.582.231.725.359.408.128 \\\\ & 48.111.745.028.410.270.193.852.110.555.964.462.294.895.493.038.196 \end{align}$$
+$$\begin{align}   & 82.148.086.513.282.306.647.093.844.609.550.582.231.725.359.408.128 \\\\
+  & 48.111.745.028.410.270.193.852.110.555.964.462.294.895.493.038.196 \end{align}$$
 
 Encontre $\sum_{n = 0, 1, \ldots, 17} {10}^n × D_{A,B}((127 + 19n) × 7^n)$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md
index 7fc0e5912cf..7cd9963812d 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md
@@ -10,7 +10,8 @@ dashedName: problem-238-infinite-string-tour
 
 Crie uma sequência de números usando o gerador de números pseudoaleatório "Blum Blum Shub":
 
-$$ s_0 = 14025256 \\\\ s_{n + 1} = {s_n}^2 \\; mod \\; 20.300.713 $$
+$$ s_0 = 14025256 \\\\
+s_{n + 1} = {s_n}^2 \\; mod \\; 20.300.713 $$
 
 Concatene esses números $s_0s_1s_2\ldots$ para criar uma string $w$ de comprimento infinito. Assim, $w = 14025256741014958470038053646\ldots$
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-240-top-dice.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-240-top-dice.md
index e8646c24acf..6adebe6a0c3 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-240-top-dice.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-240-top-dice.md
@@ -10,7 +10,9 @@ dashedName: problem-240-top-dice
 
 Há 1111 maneiras pelas quais cinco dados de 6 lados (lados numerados de 1 a 6) podem ser rolados de modo que os três maiores somem 15. Alguns exemplos:
 
-$$\begin{align} & D_1,D_2,D_3,D_4,D_5 = 4,3,6,3,5 \\\\ & D_1,D_2,D_3,D_4,D_5 = 4,3,3,5,6 \\\\ & D_1,D_2,D_3,D_4,D_5 = 3,3,3,6,6 \\\\ & D_1,D_2,D_3,D_4,D_5 = 6,6,3,3,3 \end{align}$$
+$$\begin{align}   & D_1,D_2,D_3,D_4,D_5 = 4,3,6,3,5 \\\\
+  & D_1,D_2,D_3,D_4,D_5 = 4,3,3,5,6 \\\\   & D_1,D_2,D_3,D_4,D_5 = 3,3,3,6,6 \\\\
+  & D_1,D_2,D_3,D_4,D_5 = 6,6,3,3,3 \end{align}$$
 
 De quantas maneiras vinte dados de doze lados (lados numerados de 1 a 12) podem ser rolados de modo que a soma dos dez maiores seja 70?
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-244-sliders.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-244-sliders.md
index 1696fc78feb..13098d944b5 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-244-sliders.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-244-sliders.md
@@ -16,7 +16,9 @@ Um movimento é indicado pela primeira letra maiúscula da direção (Left - Esq
 
 Para cada caminho, seu valor de verificação é calculado por (pseudocódigo):
 
-$$\begin{align} & \text{checksum} = 0 \\\\ & \text{checksum} = (\text{checksum} × 243 + m_1) \\; \text{mod} \\; 100\\,000\\,007 \\\\ & \text{checksum} = (\text{checksum} × 243 + m_2) \\; \text{mod} \\; 100\\,000\\,007 \\\\ & \ldots \\\\ & \text{checksum} = (\text{checksum} × 243 + m_n) \\; \text{mod} \\; 100\\,000\\,007 \end{align}$$
+$$\begin{align}   & \text{checksum} = 0 \\\\
+  & \text{checksum} = (\text{checksum} × 243 + m_1) \\; \text{mod} \\; 100\\,000\\,007 \\\\   & \text{checksum} = (\text{checksum} × 243 + m_2) \\; \text{mod} \\; 100\\,000\\,007 \\\\
+  & \ldots \\\\ & \text{checksum} = (\text{checksum} × 243 + m_n) \\; \text{mod} \\; 100\\,000\\,007 \end{align}$$
 
 onde $m_k$ é o valor ASCII da $k^{\text{a}}$ letra na sequência de movimento e os valores ASCII dos movimentos são:
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md
index 65d026bb0e0..1a834e20b2d 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md
@@ -16,7 +16,8 @@ Como exemplo, a imagem abaixo apresenta um conjunto de vinte pontos e alguns des
 
 Para nosso exemplo, usamos os primeiros 20 pontos ($T_{2k − 1}$, $T_{2k}$), para $k = 1, 2, \ldots, 20$, produzido com o gerador de números pseudoaleatório:
 
-$$\begin{align} S_0 & = 290.797 \\\\ S_{n+1} & = {S_n}^2 \\; \text{mod} \\; 50.515.093 \\\\ T_n & = (S_n \\; \text{mod} \\; 2000) − 1000 \end{align}$$
+$$\begin{align}   S_0 & = 290.797 \\\\
+  S_{n+1} & = {S_n}^2 \\; \text{mod} \\; 50.515.093 \\\\ T_n & = (S_n \\; \text{mod} \\; 2000) − 1000 \end{align}$$
 
 por exemplo: (527, 144), (-488, 732), (-454, − 947), …
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md
index b3caafe1d5c..ed1f2344974 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md
@@ -28,7 +28,8 @@ Como exemplo, vamos encontrar a raiz quadrada arredondada de $n = 4321$.
 
 $n$ tem 4 algarismos, então $x_0 = 7 × {10}^{\frac{4-2}{2}} = 70$.
 
-$$x_1 = \left\lfloor\frac{70 + \left\lceil\frac{4321}{70}\right\rceil}{2}\right\rfloor = 66 \\\\ x_2 = \left\lfloor\frac{66 + \left\lceil\frac{4321}{66}\right\rceil}{2}\right\rfloor = 66$$
+$$x_1 = \left\lfloor\frac{70 + \left\lceil\frac{4321}{70}\right\rceil}{2}\right\rfloor = 66 \\\\
+x_2 = \left\lfloor\frac{66 + \left\lceil\frac{4321}{66}\right\rceil}{2}\right\rfloor = 66$$
 
 Como $x_2 = x_1$, paramos aqui. Então, depois de apenas duas iterações, descobrimos que a raiz arredondada de 4321 é 66 (a raiz quadrada real é 65.7343137…).
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md
index 2f3bf6a1db4..ca53e54171e 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md
@@ -14,7 +14,9 @@ $${(k - m)}^2 + \ldots + k^2 = {(n + 1)}^2 + \ldots + {(n + m)}^2$$
 
 Alguns quadrados pivotais pequenos são
 
-$$\begin{align} & \mathbf{4}: 3^2 + \mathbf{4}^2 = 5^2 \\\\ & \mathbf{21}: {20}^2 + \mathbf{21}^2 = {29}^2 \\\\ & \mathbf{24}: {21}^2 + {22}^2 + {23}^2 + \mathbf{24}^2 = {25}^2 + {26}^2 + {27}^2 \\\\ & \mathbf{110}: {108}^2 + {109}^2 + \mathbf{110}^2 = {133}^2 + {134}^2 \\\\ \end{align}$$
+$$\begin{align}   & \mathbf{4}: 3^2 + \mathbf{4}^2 = 5^2 \\\\
+  & \mathbf{21}: {20}^2 + \mathbf{21}^2 = {29}^2 \\\\   & \mathbf{24}: {21}^2 + {22}^2 + {23}^2 + \mathbf{24}^2 = {25}^2 + {26}^2 + {27}^2 \\\\
+  & \mathbf{110}: {108}^2 + {109}^2 + \mathbf{110}^2 = {133}^2 + {134}^2 \\\\ \end{align}$$
 
 Encontre a soma de todos os quadrados pivotais distintos $≤ {10}^{10}$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md
index 51285a8de82..d3f06d8b1da 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md
@@ -10,7 +10,8 @@ dashedName: problem-282-the-ackermann-function
 
 Para números inteiros não negativos $m$, $n$, a função de Ackermann $A(m, n)$ é definida da seguinte forma:
 
-$$A(m, n) = \begin{cases} n + 1                 & \text{if $m = 0$}             \\\\ A(m - 1, 1)           & \text{if $m > 0$ and $n = 0$} \\\\ A(m - 1, A(m, n - 1)) & \text{if $m > 0$ and $n > 0$} \end{cases}$$
+$$A(m, n) = \begin{cases} n + 1                 & \text{if $m = 0$}             \\\\
+A(m - 1, 1)           & \text{if $m > 0$ and $n = 0$} \\\\ A(m - 1, A(m, n - 1)) & \text{if $m > 0$ and $n > 0$} \end{cases}$$
 
 Por exemplo $A(1, 0) = 2$, $A(2, 2) = 7$ e $A(3, 4) = 125$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md
index 9877641aeb5..f420fedef90 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md
@@ -10,7 +10,8 @@ dashedName: problem-288-an-enormous-factorial
 
 Para qualquer número primo $p$, o número $N(p,q)$ é definido por $N(p,q) = \sum_{n=0}^q T_n \times p^n$ com $T_n$ gerado pelo seguinte gerador aleatório de números:
 
-$$\begin{align} & S_0 = 290797 \\\\ & S_{n + 1} = {S_n}^2\bmod 50.515.093 \\\\ & T_n = S_n\bmod p \end{align}$$
+$$\begin{align}   & S_0 = 290797 \\\\
+  & S_{n + 1} = {S_n}^2\bmod 50.515.093 \\\\ & T_n = S_n\bmod p \end{align}$$
 
 Considere $Nfac(p,q)$ como o fatorial de $N(p,q)$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md
index 1ee7d18c3bd..5b78321cfae 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md
@@ -16,7 +16,8 @@ Chamamos a área convexa criada no cruzamento de dois círculos de um orifício
 
 Considere os círculos:
 
-$$\begin{align} & C_0: x^2 + y^2 = 25 \\\\ & C_1: {(x + 4)}^2 + {(y - 4)}^2 = 1 \\\\ & C_2: {(x - 12)}^2 + {(y - 4)}^2 = 65 \end{align}$$
+$$\begin{align}   & C_0: x^2 + y^2 = 25 \\\\
+  & C_1: {(x + 4)}^2 + {(y - 4)}^2 = 1 \\\\ & C_2: {(x - 12)}^2 + {(y - 4)}^2 = 65 \end{align}$$
 
 Os círculos $C_0$, $C_1$ e $C_2$ estão desenhados na imagem abaixo.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md
index 6fbf754ad7a..6fd80ed9cae 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md
@@ -19,7 +19,9 @@ Considere $C(n)$ como o número de ciclos que passam exatamente uma vez por todo
 
 Também pode ser verificado que:
 
-$$\begin{align} & C(1) = C(2) = 1 \\\\ & C(5) = 71.328.803.586.048 \\\\ & C(10 000)\bmod {10}^8 = 37.652.224 \\\\ & C(10 000)\bmod {13}^8 = 617.720.485 \\\\ \end{align}$$
+$$\begin{align}   & C(1) = C(2) = 1 \\\\
+  & C(5) = 71.328.803.586.048 \\\\   & C(10 000)\bmod {10}^8 = 37.652.224 \\\\
+  & C(10 000)\bmod {13}^8 = 617.720.485 \\\\ \end{align}$$
 
 Encontre $C(C(C(10.000)))\bmod {13}^8$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
index 7fed0ec8d22..892ae80c9bf 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
@@ -12,7 +12,11 @@ Considere o número real $\sqrt{2} + \sqrt{3}$.
 
 Quando calculamos as potências pares de $\sqrt{2} + \sqrt{3}$ obtemos:
 
-$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
+$$\begin{align}   & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\
+  & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\   & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\
+  & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\   & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\
+  & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\   & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\
+  & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
 
 Parece que o número de noves consecutivos no início da parte fracionária dessas potências não diminui. Na verdade, pode ser provado que a parte fracionária de ${(\sqrt{2} + \sqrt{3})}^{2n}$ aproxima-se de 1 para $n$ grandes.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md
index 50afc3121a3..23509334be9 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md
@@ -12,7 +12,9 @@ Considere $f(n)$ como o número de maneiras que se pode preencher uma torre $3×
 
 Por exemplo (com $q = 100.000.007$):
 
-$$\begin{align} & f(2) = 229, \\\\ & f(4) = 117.805, \\\\ & f(10)\bmod q = 96.149.360, \\\\ & f({10}^3)\bmod q = 24.806.056, \\\\ & f({10}^6)\bmod q = 30.808.124. \end{align}$$
+$$\begin{align}   & f(2) = 229, \\\\
+  & f(4) = 117.805, \\\\   & f(10)\bmod q = 96.149.360, \\\\
+  & f({10}^3)\bmod q = 24.806.056, \\\\ & f({10}^6)\bmod q = 30.808.124. \end{align}$$
 
 Encontre $f({10}^{10000})\bmod 100.000.007$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md
index 4d74896f27f..7d42d0c9001 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md
@@ -10,11 +10,13 @@ dashedName: problem-330-eulers-number
 
 Uma sequência infinita de números reais $a(n)$ é definida para todos os números inteiros $n$ da seguinte forma:
 
-$$ a(n) = \begin{cases} 1                                                       & n < 0 \\\\ \displaystyle \sum_{i = 1}^{\infty} \frac{a(n - 1)}{i!} & n \ge 0 \end{cases} $$
+$$ a(n) = \begin{cases} 1                                                       & n < 0 \\\\
+\displaystyle \sum_{i = 1}^{\infty} \frac{a(n - 1)}{i!} & n \ge 0 \end{cases} $$
 
 Por exemplo:
 
-$$\begin{align} & a(0) = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = e − 1 \\\\ & a(1) = \frac{e − 1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = 2e − 3 \\\\ & a(2) = \frac{2e − 3}{1!} + \frac{e − 1}{2!} + \frac{1}{3!} + \ldots = \frac{7}{2} e − 6 \end{align}$$
+$$\begin{align}   & a(0) = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = e − 1 \\\\
+  & a(1) = \frac{e − 1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = 2e − 3 \\\\ & a(2) = \frac{2e − 3}{1!} + \frac{e − 1}{2!} + \frac{1}{3!} + \ldots = \frac{7}{2} e − 6 \end{align}$$
 
 com $e = 2.7182818\ldots$ sendo a constante de Euler.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md
index 216034dba9f..7c90a2dbc00 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md
@@ -14,7 +14,8 @@ Consideremos apenas aquelas partições em que nenhum dos termos pode dividir qu
 
 Muitos números inteiros têm mais de uma partição válida, sendo o primeiro 11 tendo as duas partições que seguem.
 
-$$\begin{align} & 11 = 2 + 9 = (2^1 \times 3^0 + 2^0 \times 3^2) \\\\ & 11 = 8 + 3 = (2^3 \times 3^0 + 2^0 \times 3^1) \end{align}$$
+$$\begin{align}   & 11 = 2 + 9 = (2^1 \times 3^0 + 2^0 \times 3^2) \\\\
+  & 11 = 8 + 3 = (2^3 \times 3^0 + 2^0 \times 3^1) \end{align}$$
 
 Vamos definir $P(n)$ como o número de partições válidas de $n$. Por exemplo, $P(11) = 2$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md
index 4492cb51bd6..4d9a44c89c3 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md
@@ -16,7 +16,10 @@ Por exemplo, considere duas tigelas adjacentes contendo 2 e 3 feijões, respecti
 
 Você recebe as seguintes sequências:
 
-$$\begin{align} & t_0 = 123456, \\\\ & t_i = \begin{cases} \frac{t_{i - 1}}{2},               & \text{if $t_{i - 1}$ é par} \\\\ \left\lfloor\frac{t_{i - 1}}{2}\right\rfloor \oplus 926252, & \text{if $t_{i - 1}$ é ímpar} \end{cases} \\\\ & \qquad \text{onde $⌊x⌋$ é a função piso e $\oplus$ é o operador bitwise XOR.} \\\\ & b_i = (t_i\bmod 2^{11}) + 1. \end{align}$$
+$$\begin{align}   & t_0 = 123456, \\\\
+  & t_i = \begin{cases}          \frac{t_{i - 1}}{2},               & \text{if $t_{i - 1}$ é par} \\\\
+         \left\lfloor\frac{t_{i - 1}}{2}\right\rfloor \oplus 926252, & \text{if $t_{i - 1}$ é ímpar}          \end{cases} \\\\
+         & \qquad \text{onde $⌊x⌋$ é a função piso e $\oplus$ é o operador bitwise XOR.} \\\\ & b_i = (t_i\bmod 2^{11}) + 1. \end{align}$$
 
 Os dois primeiros termos da última sequência são $b_1 = 289$ e $b_2 = 145$. Se começarmos com $b_1$ e $b_2$ feijões em duas tigelas adjacentes, 3419100 movimentos seriam necessários para terminar o jogo.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md
index 9d117967a57..c81525c9424 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md
@@ -10,7 +10,8 @@ dashedName: problem-340-crazy-function
 
 Para números inteiros fixos $a$, $b$, $c$, defina a função maluca $F(n)$ da seguinte forma:
 
-$$\begin{align} & F(n) = n - c \\;\text{ para todo } n > b \\\\ & F(n) = F(a + F(a + F(a + F(a + n)))) \\;\text{ para todo } n ≤ b. \end{align}$$
+$$\begin{align}   & F(n) = n - c \\;\text{ para todo } n > b \\\\
+  & F(n) = F(a + F(a + F(a + F(a + n)))) \\;\text{ para todo } n ≤ b. \end{align}$$
 
 Além disso, defina $S(a, b, c) = \displaystyle\sum_{n = 0}^b F(n)$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md
index 7d7705ce068..e4ba716c2b7 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md
@@ -10,7 +10,8 @@ dashedName: problem-341-golombs-self-describing-sequence
 
 A sequência autodescritiva de Golomb ($G(n)$) é a única sequência não decrescente de números naturais, tal que $n$ aparece exatamente $G(n)$ vezes na sequência. Os valores de $G(n)$ para os primeiros $n$ são
 
-$$\begin{array}{c} n    & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & \ldots \\\\ G(n) & 1 & 2 & 2 & 3 & 3 & 4 & 4 & 4 & 5 & 5  &  5 &  6 &  6 &  6 &  6 & \ldots \end{array}$$
+$$\begin{array}{c}   n    & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & \ldots \\\\
+  G(n) & 1 & 2 & 2 & 3 & 3 & 4 & 4 & 4 & 5 & 5  &  5 &  6 &  6 &  6 &  6 & \ldots \end{array}$$
 
 Você é informado de que $G({10}^3) = 86$, $G({10}^6) = 6137$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md
index 5603798be3b..ba2d281cb26 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md
@@ -12,11 +12,20 @@ Definimos a soma interna da matriz como a soma máxima dos elementos da matriz c
 
 Por exemplo, a soma interna da matriz abaixo é igual a $3315 ( = 863 + 383 + 343 + 959 + 767)$:
 
-$$\begin{array}{rrrrr} 7                &  53                & 183                & 439                & \color{lime}{863} \\\\ 497                & \color{lime}{383} & 563                &  79                & 973 \\\\ 287                &  63                & \color{lime}{343} & 169                & 583 \\\\ 627                & 343                & 773                & \color{lime}{959} & 943 \\\\ \color{lime}{767} & 473                & 103                & 699                & 303 \end{array}$$
+$$\begin{array}{rrrrr}     7                &  53                & 183                & 439                & \color{lime}{863} \\\\
+  497                & \color{lime}{383} & 563                &  79                & 973 \\\\   287                &  63                & \color{lime}{343} & 169                & 583 \\\\
+  627                & 343                & 773                & \color{lime}{959} & 943 \\\\ \color{lime}{767} & 473                & 103                & 699                & 303 \end{array}$$
 
 Encontre a soma interna da matriz de:
 
-$$\\begin{array}{r} 7 &  53 & 183 & 439 & 863 & 497 & 383 & 563 &  79 & 973 & 287 &  63 & 343 & 169 & 583 \\\\ 627 & 343 & 773 & 959 & 943 & 767 & 473 & 103 & 699 & 303 & 957 & 703 & 583 & 639 & 913 \\\\ 447 & 283 & 463 &  29 &  23 & 487 & 463 & 993 & 119 & 883 & 327 & 493 & 423 & 159 & 743 \\\\ 217 & 623 &   3 & 399 & 853 & 407 & 103 & 983 &  89 & 463 & 290 & 516 & 212 & 462 & 350 \\\\ 960 & 376 & 682 & 962 & 300 & 780 & 486 & 502 & 912 & 800 & 250 & 346 & 172 & 812 & 350 \\\\ 870 & 456 & 192 & 162 & 593 & 473 & 915 &  45 & 989 & 873 & 823 & 965 & 425 & 329 & 803 \\\\ 973 & 965 & 905 & 919 & 133 & 673 & 665 & 235 & 509 & 613 & 673 & 815 & 165 & 992 & 326 \\\\ 322 & 148 & 972 & 962 & 286 & 255 & 941 & 541 & 265 & 323 & 925 & 281 & 601 &  95 & 973 \\\\ 445 & 721 &  11 & 525 & 473 &  65 & 511 & 164 & 138 & 672 &  18 & 428 & 154 & 448 & 848 \\\\ 414 & 456 & 310 & 312 & 798 & 104 & 566 & 520 & 302 & 248 & 694 & 976 & 430 & 392 & 198 \\\\ 184 & 829 & 373 & 181 & 631 & 101 & 969 & 613 & 840 & 740 & 778 & 458 & 284 & 760 & 390 \\\\ 821 & 461 & 843 & 513 &  17 & 901 & 711 & 993 & 293 & 157 & 274 &  94 & 192 & 156 & 574 \\\\ 34 & 124 &   4 & 878 & 450 & 476 & 712 & 914 & 838 & 669 & 875 & 299 & 823 & 329 & 699 \\\\ 815 & 559 & 813 & 459 & 522 & 788 & 168 & 586 & 966 & 232 & 308 & 833 & 251 & 631 & 107 \\\\ 813 & 883 & 451 & 509 & 615 &  77 & 281 & 613 & 459 & 205 & 380 & 274 & 302 &  35 & 805 \end{array}$$
+$$\\begin{array}{r}     7 &  53 & 183 & 439 & 863 & 497 & 383 & 563 &  79 & 973 & 287 &  63 & 343 & 169 & 583 \\\\
+  627 & 343 & 773 & 959 & 943 & 767 & 473 & 103 & 699 & 303 & 957 & 703 & 583 & 639 & 913 \\\\   447 & 283 & 463 &  29 &  23 & 487 & 463 & 993 & 119 & 883 & 327 & 493 & 423 & 159 & 743 \\\\
+  217 & 623 &   3 & 399 & 853 & 407 & 103 & 983 &  89 & 463 & 290 & 516 & 212 & 462 & 350 \\\\   960 & 376 & 682 & 962 & 300 & 780 & 486 & 502 & 912 & 800 & 250 & 346 & 172 & 812 & 350 \\\\
+  870 & 456 & 192 & 162 & 593 & 473 & 915 &  45 & 989 & 873 & 823 & 965 & 425 & 329 & 803 \\\\   973 & 965 & 905 & 919 & 133 & 673 & 665 & 235 & 509 & 613 & 673 & 815 & 165 & 992 & 326 \\\\
+  322 & 148 & 972 & 962 & 286 & 255 & 941 & 541 & 265 & 323 & 925 & 281 & 601 &  95 & 973 \\\\   445 & 721 &  11 & 525 & 473 &  65 & 511 & 164 & 138 & 672 &  18 & 428 & 154 & 448 & 848 \\\\
+  414 & 456 & 310 & 312 & 798 & 104 & 566 & 520 & 302 & 248 & 694 & 976 & 430 & 392 & 198 \\\\   184 & 829 & 373 & 181 & 631 & 101 & 969 & 613 & 840 & 740 & 778 & 458 & 284 & 760 & 390 \\\\
+  821 & 461 & 843 & 513 &  17 & 901 & 711 & 993 & 293 & 157 & 274 &  94 & 192 & 156 & 574 \\\\    34 & 124 &   4 & 878 & 450 & 476 & 712 & 914 & 838 & 669 & 875 & 299 & 823 & 329 & 699 \\\\
+  815 & 559 & 813 & 459 & 522 & 788 & 168 & 586 & 966 & 232 & 308 & 833 & 251 & 631 & 107 \\\\ 813 & 883 & 451 & 509 & 615 &  77 & 281 & 613 & 459 & 205 & 380 & 274 & 302 &  35 & 805 \end{array}$$
 
 # --hints--
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md
index 98b506b5d96..b413a309e08 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md
@@ -14,7 +14,9 @@ Considere os números palíndromos que podem ser expressos como a soma de um qua
 
 Por exemplo, 5229225 é um número palíndromo e pode ser expresso de exatamente 4 formas diferentes:
 
-$$\begin{align} & {2285}^2 + {20}^3 \\\\ & {2223}^2 + {66}^3 \\\\ & {1810}^2 + {125}^3 \\\\ & {1197}^2 + {156}^3 \end{align}$$
+$$\begin{align}   & {2285}^2 + {20}^3 \\\\
+  & {2223}^2 + {66}^3 \\\\   & {1810}^2 + {125}^3 \\\\
+  & {1197}^2 + {156}^3 \end{align}$$
 
 Encontre a soma dos cinco menores números palíndromos deste tipo.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md
index f59e83381f4..f84943ffde1 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md
@@ -16,7 +16,9 @@ O mínimo múltiplo comum, ou $lcm$, de uma lista é o menor número natural div
 
 Considere $f(G, L, N)$ como o número de listas de tamanho $N$ com $gcd ≥ G$ e $lcm ≤ L$. Por exemplo:
 
-$$\begin{align} & f(10, 100, 1) = 91 \\\\ & f(10, 100, 2) = 327 \\\\ & f(10, 100, 3) = 1135 \\\\ & f(10, 100, 1000)\bmod {101}^4 = 3.286.053 \end{align}$$
+$$\begin{align}   & f(10, 100, 1) = 91 \\\\
+  & f(10, 100, 2) = 327 \\\\   & f(10, 100, 3) = 1135 \\\\
+  & f(10, 100, 1000)\bmod {101}^4 = 3.286.053 \end{align}$$
 
 Encontre $f({10}^6, {10}^{12}, {10}^{18})\bmod {101}^4$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md
index 56dfdb742f1..2c0aee9cc50 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md
@@ -14,11 +14,16 @@ Quando é multiplicado por 1, 2, 3, 4, ... $n$, todos os produtos têm exatament
 
 O menor número cíclico é o número de 6 algarismos 142857:
 
-$$\begin{align} & 142857 × 1 = 142857 \\\\ & 142857 × 2 = 285714 \\\\ & 142857 × 3 = 428571 \\\\ & 142857 × 4 = 571428 \\\\ & 142857 × 5 = 714285 \\\\ & 142857 × 6 = 857142 \end{align}$$
+$$\begin{align}   & 142857 × 1 = 142857 \\\\
+  & 142857 × 2 = 285714 \\\\   & 142857 × 3 = 428571 \\\\
+  & 142857 × 4 = 571428 \\\\   & 142857 × 5 = 714285 \\\\
+  & 142857 × 6 = 857142 \end{align}$$
 
 O próximo número cíclico é 0588235294117647, com 16 algarismos:
 
-$$\begin{align} & 0588235294117647 × 1 = 0588235294117647 \\\\ & 0588235294117647 × 2 = 1176470588235294 \\\\ & 0588235294117647 × 3 = 1764705882352941 \\\\ & \ldots \\\\ & 0588235294117647 × 16 = 9411764705882352 \end{align}$$
+$$\begin{align}   & 0588235294117647 × 1 = 0588235294117647 \\\\
+  & 0588235294117647 × 2 = 1176470588235294 \\\\   & 0588235294117647 × 3 = 1764705882352941 \\\\
+  & \ldots \\\\ & 0588235294117647 × 16 = 9411764705882352 \end{align}$$
 
 Observe que, para números cíclicos, zeros à esquerda são importantes.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md
index 09bd7a699f0..b44d7798279 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md
@@ -27,7 +27,10 @@ No fim, cada pessoa na fila pegará um quarto no hotel.
 
 Defina $P(f, r)$ como $n$ se a pessoa $n$ ocupar o quarto $r$ no andar $f$, e 0 se ninguém ocupar o quarto. Aqui estão alguns exemplos:
 
-$$\begin{align} & P(1, 1) = 1 \\\\ & P(1, 2) = 3 \\\\ & P(2, 1) = 2 \\\\ & P(10, 20) = 440 \\\\ & P(25, 75) = 4863 \\\\ & P(99, 100) = 19454 \end{align}$$
+$$\begin{align}   & P(1, 1) = 1 \\\\
+  & P(1, 2) = 3 \\\\   & P(2, 1) = 2 \\\\
+  & P(10, 20) = 440 \\\\   & P(25, 75) = 4863 \\\\
+  & P(99, 100) = 19454 \end{align}$$
 
 Encontre a soma de todos os $P(f, r)$ para todos os números positivos $f$ e $r$, tal que $f × r = 71.328.803.586.048$ e dê os últimos 8 algarismos como resposta.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md
index c784f900c1a..23b0b04003f 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md
@@ -20,7 +20,8 @@ Definimos $\\{A_n\\}$ como uma sequência ordenada de inteiros, de forma que a e
 
 Os primeiros termos de $A_n$ são atribuídos da seguinte forma:
 
-$$\begin{array}{cr} n   & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 &  8 &  9 & 10 & 11 & 12 & \ldots \\\\ A_n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 9 & 10 & 11 & 12 & 13 & 18 & \ldots \end{array}$$
+$$\begin{array}{cr}   n   & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 &  8 &  9 & 10 & 11 & 12 & \ldots \\\\
+  A_n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 9 & 10 & 11 & 12 & 13 & 18 & \ldots \end{array}$$
 
 Também podemos verificar que $A_{100} = 3251$ e $A_{1000} = 80.852.364.498$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md
index cebfef535d9..bdca3715e31 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md
@@ -16,7 +16,8 @@ Consideremos agora outra série harmônica modificada, omitindo da série harmô
 
 Estes 20 termos omitidos são:
 
-$$\dfrac{1}{111}, \dfrac{1}{222}, \dfrac{1}{333}, \dfrac{1}{444}, \dfrac{1}{555}, \dfrac{1}{666}, \dfrac{1}{777}, \dfrac{1}{888}, \dfrac{1}{999}, \dfrac{1}{1000}, \dfrac{1}{1110}, \\\\ \dfrac{1}{1111}, \dfrac{1}{1112}, \dfrac{1}{1113}, \dfrac{1}{1114}, \dfrac{1}{1115}, \dfrac{1}{1116}, \dfrac{1}{1117}, \dfrac{1}{1118}, \dfrac{1}{1119}$$
+$$\dfrac{1}{111}, \dfrac{1}{222}, \dfrac{1}{333}, \dfrac{1}{444}, \dfrac{1}{555}, \dfrac{1}{666}, \dfrac{1}{777}, \dfrac{1}{888}, \dfrac{1}{999}, \dfrac{1}{1000}, \dfrac{1}{1110}, \\\\
+\dfrac{1}{1111}, \dfrac{1}{1112}, \dfrac{1}{1113}, \dfrac{1}{1114}, \dfrac{1}{1115}, \dfrac{1}{1116}, \dfrac{1}{1117}, \dfrac{1}{1118}, \dfrac{1}{1119}$$
 
 Esta série também converge.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md
index 49589b217ae..0f96415a429 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md
@@ -10,7 +10,8 @@ dashedName: problem-375-minimum-of-subsequences
 
 Considere $S_n$ como uma sequência de números inteiros produzida com o seguinte gerador de números pseudoaleatórios:
 
-$$\begin{align} S_0 & = 290.797 \\\\ S_{n + 1} & = {S_n}^2\bmod 50.515.093 \end{align}$$
+$$\begin{align}         S_0 & = 290.797 \\\\
+  S_{n + 1} & = {S_n}^2\bmod 50.515.093 \end{align}$$
 
 Considere $A(i, j)$ como o mínimo dos números $S_i, S_{i + 1}, \ldots, S_j$ para $i ≤ j$. Considere $M(N) = \sum A(i, j)$ para $1 ≤ i ≤ j ≤ N$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md
index 89aaca20743..65bf18c77af 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md
@@ -10,7 +10,9 @@ dashedName: problem-376-nontransitive-sets-of-dice
 
 Considere o seguinte conjunto de dados com valores fora do padrão de 1 a 6:
 
-$$\begin{array}{} \text{Die A: } & 1 & 4 & 4 & 4 & 4 & 4 \\\\ \text{Die B: } & 2 & 2 & 2 & 5 & 5 & 5 \\\\ \text{Die C: } & 3 & 3 & 3 & 3 & 3 & 6 \\\\ \end{array}$$
+$$\begin{array}{}   \text{Die A: } & 1 & 4 & 4 & 4 & 4 & 4 \\\\
+  \text{Die B: } & 2 & 2 & 2 & 5 & 5 & 5 \\\\   \text{Die C: } & 3 & 3 & 3 & 3 & 3 & 6 \\\\
+\end{array}$$
 
 Um jogo é disputado por dois jogadores que escolhem um dado por vez e o rolam. O jogador que rolar nos dados o maior valor ganha.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md
index 41615e1b037..bcc15e745d6 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md
@@ -10,7 +10,9 @@ dashedName: problem-38-pandigital-multiples
 
 Pegue o número 192 e multiplique-o por cada um entre 1, 2 e 3:
 
-$$\begin{align} 192 × 1 = 192\\\\ 192 × 2 = 384\\\\ 192 × 3 = 576\\\\ \end{align}$$
+$$\begin{align}   192 × 1 = 192\\\\
+  192 × 2 = 384\\\\   192 × 3 = 576\\\\
+\end{align}$$
 
 Ao concatenar cada produto, chegamos ao total 192384576. Esse resultado possui 9 algarismos e usa todos os números de 1 a 9 pelo menos uma vez. Chamaremos 192384576 o produto concatenado de 192 e (1, 2, 3).
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md
index eeb08669331..ca19ed14714 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md
@@ -18,7 +18,9 @@ Além disso, considere a sequência somatória de $b(n)$: $s(n) = \displaystyle\
 
 Os primeiros valores destas sequências são:
 
-$$\begin{array}{lr} n    & 0 & 1 & 2 &  3 & 4 & 5 &  6 & 7 \\\\ a(n) & 0 & 0 & 0 &  1 & 0 & 0 &  1 & 2 \\\\ b(n) & 1 & 1 & 1 & -1 & 1 & 1 & -1 & 1 \\\\ s(n) & 1 & 2 & 3 &  2 & 3 & 4 &  3 & 4 \end{array}$$
+$$\begin{array}{lr}   n    & 0 & 1 & 2 &  3 & 4 & 5 &  6 & 7 \\\\
+  a(n) & 0 & 0 & 0 &  1 & 0 & 0 &  1 & 2 \\\\   b(n) & 1 & 1 & 1 & -1 & 1 & 1 & -1 & 1 \\\\
+  s(n) & 1 & 2 & 3 &  2 & 3 & 4 &  3 & 4 \end{array}$$
 
 A sequência $s(n)$ tem a incrível propriedade de que todos os elementos são positivos e de que todo número inteiro positivo $k$ ocorre exatamente $k$ vezes.
 
@@ -28,7 +30,8 @@ Ex.: $g(3, 3) = 6$, $g(4, 2) = 7$ e $g(54321, 12345) = 1.220.847.710$.
 
 Considere $F(n)$ como a sequência de Fibonacci definida por:
 
-$$\begin{align} & F(0) = F(1) = 1 \text{ e} \\\\ & F(n) = F(n - 1) + F(n - 2) \text{ para } n > 1. \end{align}$$
+$$\begin{align}   & F(0) = F(1) = 1 \text{ e} \\\\
+  & F(n) = F(n - 1) + F(n - 2) \text{ para } n > 1. \end{align}$$
 
 Defina $GF(t) = g(F(t), F(t - 1))$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md
index 9cee9f05ebd..297d71ca0e4 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md
@@ -32,7 +32,9 @@ Considere $C(n, a, b)$ como o pior caso de custo obtido por uma estratégia idea
 
 Aqui estão alguns exemplos:
 
-$$\begin{align} & C(5, 2, 3) = 5 \\\\ & C(500, \sqrt{2}, \sqrt{3}) = 13.220\\,731\\,97\ldots \\\\ & C(20.000, 5, 7) = 82 \\\\ & C(2.000.000, √5, √7) = 49.637\\,559\\,55\ldots \\\\ \end{align}$$
+$$\begin{align}   & C(5, 2, 3) = 5 \\\\
+  & C(500, \sqrt{2}, \sqrt{3}) = 13.220\\,731\\,97\ldots \\\\   & C(20.000, 5, 7) = 82 \\\\
+  & C(2.000.000, √5, √7) = 49.637\\,559\\,55\ldots \\\\ \end{align}$$
 
 Considere $F_k$ como sendo os números de Fibonacci: $F_k = F_{k - 1} + F_{k - 2}$ com casos base $F_1 = F_2 = 1$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md
index 4b69cdbaccf..25f6f2fec46 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md
@@ -16,7 +16,8 @@ Isso também funciona com números que têm menos de 4 algarismos se colocarmos
 
 Ex: vamos começar com o número 0837:
 
-$$\begin{align} & 8730 - 0378 = 8352 \\\\ & 8532 - 2358 = 6174 \end{align}$$
+$$\begin{align}   & 8730 - 0378 = 8352 \\\\
+  & 8532 - 2358 = 6174 \end{align}$$
 
 6174 é chamado de constante de Kaprekar. O processo de ordenar e subtrair e repetir isso até chegar a 0 ou à constante de Kaprekar é chamado de rotina de Kaprekar.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md
index 584a923a416..40f6ca788ef 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md
@@ -10,7 +10,12 @@ dashedName: problem-417-reciprocal-cycles-ii
 
 Em uma fração unitária, o numerador é 1. A representação decimal das frações unitárias com denominadores de 2 a 10 é a seguinte:
 
-$$\begin{align} & \frac{1}{2}  = 0.5 \\\\ & \frac{1}{3}  = 0.(3) \\\\ & \frac{1}{4}  = 0.25 \\\\ & \frac{1}{5}  = 0.2 \\\\ & \frac{1}{6}  = 0.1(6) \\\\ & \frac{1}{7}  = 0.(142857) \\\\ & \frac{1}{8}  = 0.125 \\\\ & \frac{1}{9}  = 0.(1) \\\\ & \frac{1}{10} = 0.1 \\\\ \end{align}$$
+$$\begin{align}   & \frac{1}{2}  = 0.5 \\\\
+  & \frac{1}{3}  = 0.(3) \\\\   & \frac{1}{4}  = 0.25 \\\\
+  & \frac{1}{5}  = 0.2 \\\\   & \frac{1}{6}  = 0.1(6) \\\\
+  & \frac{1}{7}  = 0.(142857) \\\\   & \frac{1}{8}  = 0.125 \\\\
+  & \frac{1}{9}  = 0.(1) \\\\   & \frac{1}{10} = 0.1 \\\\
+\end{align}$$
 
 A expressão $0.1(6)$ significa $0.16666666\dots$, e tem um ciclo recorrente de 1 algarismo. Pode ser visto que $\frac{1}{7}$ tem um ciclo recorrente de 6 algarismos.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md
index a3787a24298..58838c46d23 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md
@@ -12,11 +12,14 @@ Uma matriz de números inteiros positivos é uma matriz cujos elementos são tod
 
 Algumas matrizes de números inteiros positivos podem ser expressas como um quadrado de uma matriz de números inteiros positivos de duas formas diferentes. Exemplo:
 
-$$\begin{pmatrix} 40 & 12 \\\\ 48 & 40 \end{pmatrix} =
+$$\begin{pmatrix}   40 & 12 \\\\
+  48 & 40 \end{pmatrix} =
 {\begin{pmatrix}
-  2 & 3 \\\\ 12 & 2 \end{pmatrix}}^2 =
+  2 & 3 \\\\
+ 12 & 2 \end{pmatrix}}^2 =
 {\begin{pmatrix}
-  6 & 1 \\\\ 4 & 6 \end{pmatrix}}^2$$
+  6 & 1 \\\\
+  4 & 6 \end{pmatrix}}^2$$
 
 Definimos $F(N)$ como a quantidade de matrizes de números inteiros positivos 2x2 que têm um traço inferior a N e que podem ser expressas como um quadrado de uma matriz de números inteiros positivos de duas formas diferentes.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md
index 954ec292c8b..158605d0723 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md
@@ -14,7 +14,8 @@ Um dado de 6 lados é lançado $n$ vezes. Considere $c$ como o número de pares
 
 Por exemplo, se $n = 7$ e os valores dos lançamentos dos dados são (1, 1, 5, 6, 6, 6, 3), os seguintes pares de lançamentos consecutivos dão o mesmo valor:
 
-$$\begin{align} & (\underline{1}, \underline{1}, 5, 6, 6, 6, 3) \\\\ & (1, 1, 5, \underline{6}, \underline{6}, 6, 3) \\\\ & (1, 1, 5, 6, \underline{6}, \underline{6}, 3) \end{align}$$
+$$\begin{align}   & (\underline{1}, \underline{1}, 5, 6, 6, 6, 3) \\\\
+  & (1, 1, 5, \underline{6}, \underline{6}, 6, 3) \\\\ & (1, 1, 5, 6, \underline{6}, \underline{6}, 3) \end{align}$$
 
 Portanto, $c = 3$ para (1, 1, 5, 6, 6, 6, 3).
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md
index 034d2650cbb..e0d7390e6a6 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md
@@ -24,7 +24,8 @@ Pode-se mostrar que após um número suficiente de movimentos, o sistema evolui
 
 Definimos a sequência $\\{t_i\\}$:
 
-$$\begin{align} & s_0 = 290.797 \\\\ & s_{k + 1} = {s_k}^2\bmod 50.515.093 \\\\ & t_k = (s_k\bmod 64) + 1 \end{align}$$
+$$\begin{align}   & s_0 = 290.797 \\\\
+  & s_{k + 1} = {s_k}^2\bmod 50.515.093 \\\\ & t_k = (s_k\bmod 64) + 1 \end{align}$$
 
 Começando da configuração inicial $(t_0, t_1, \ldots, t_{10})$, o estado final se torna [1, 3, 10, 24, 51, 75].
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md
index 123b57bc3a1..2947d7890c4 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md
@@ -10,7 +10,8 @@ dashedName: problem-433-steps-in-euclids-algorithm
 
 Considere $E(x_0, y_0)$ como o número de etapas necessárias para determinar o máximo divisor comum de $x_0$ e $y_0$ com o algoritmo de Euclides. Mais formalmente:
 
-$$\begin{align} & x_1 = y_0, y_1 = x_0\bmod y_0 \\\\ & x_n = y_{n - 1}, y_n = x_{n - 1}\bmod y_{n - 1} \end{align}$$
+$$\begin{align}   & x_1 = y_0, y_1 = x_0\bmod y_0 \\\\
+  & x_n = y_{n - 1}, y_n = x_{n - 1}\bmod y_{n - 1} \end{align}$$
 
 $E(x_0, y_0)$ é o menor $n$, tal que $y_n = 0$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md
index ac5a1c3d94d..532ca109f40 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md
@@ -16,7 +16,11 @@ Mas há mais:
 
 Se olharmos mais de perto:
 
-$$\begin{align} & 1 + 8 = 9 \\\\ & 8 + 9 = 17 ≡ 6\bmod 11 \\\\ & 9 + 6 = 15 ≡ 4\bmod 11 \\\\ & 6 + 4 = 10 \\\\ & 4 + 10 = 14 ≡ 3\bmod 11 \\\\ & 10 + 3 = 13 ≡ 2\bmod 11 \\\\ & 3 + 2 = 5 \\\\ & 2 + 5 = 7 \\\\ & 5 + 7 = 12 ≡ 1\bmod 11. \end{align}$$
+$$\begin{align}   & 1 + 8 = 9 \\\\
+  & 8 + 9 = 17 ≡ 6\bmod 11 \\\\   & 9 + 6 = 15 ≡ 4\bmod 11 \\\\
+  & 6 + 4 = 10 \\\\   & 4 + 10 = 14 ≡ 3\bmod 11 \\\\
+  & 10 + 3 = 13 ≡ 2\bmod 11 \\\\   & 3 + 2 = 5 \\\\
+  & 2 + 5 = 7 \\\\ & 5 + 7 = 12 ≡ 1\bmod 11. \end{align}$$
 
 Portanto, as potências de 8 mod 11 são cíclicas com o período de 10 e $8^n + 8^{n + 1} ≡ 8^{n + 2} (\text{mod } 11)$. 8 é chamado de raiz primitiva de Fibonacci de 11.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md
index bb73b3694c6..ad47be0e53b 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md
@@ -24,7 +24,8 @@ Considere $S(L)$ como a soma tripla $\sum_{a, b, c} gcd(T(c^a), T(c^b))$ para $1
 
 Por exemplo:
 
-$$\begin{align} & S(2) = 10.444 \\\\ & S(3) = 1.292.115.238.446.807.016.106.539.989 \\\\ & S(4)\bmod 987.898.789 = 670.616.280. \end{align}$$
+$$\begin{align}   & S(2) = 10.444 \\\\
+  & S(3) = 1.292.115.238.446.807.016.106.539.989 \\\\ & S(4)\bmod 987.898.789 = 670.616.280. \end{align}$$
 
 Encontre $S(2000)\bmod 987.898.789$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md
index 170fc482c67..cd183af3e3e 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md
@@ -10,11 +10,13 @@ dashedName: problem-443-gcd-sequence
 
 Considere $g(n)$ como uma sequência definida assim:
 
-$$\begin{align} & g(4) = 13, \\\\ & g(n) = g(n-1) + gcd(n, g(n - 1)) \text{ para } n > 4. \end{align}$$
+$$\begin{align}   & g(4) = 13, \\\\
+  & g(n) = g(n-1) + gcd(n, g(n - 1)) \text{ para } n > 4. \end{align}$$
 
 Seus primeiros valores são:
 
-$$\begin{array}{l} n    & 4  & 5  & 6  & 7  & 8  & 9  & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & \ldots \\\\ g(n) & 13 & 14 & 16 & 17 & 18 & 27 & 28 & 29 & 30 & 31 & 32 & 33 & 34 & 51 & 54 & 55 & 60 & \ldots \end{array}$$
+$$\begin{array}{l}   n    & 4  & 5  & 6  & 7  & 8  & 9  & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & \ldots \\\\
+  g(n) & 13 & 14 & 16 & 17 & 18 & 27 & 28 & 29 & 30 & 31 & 32 & 33 & 34 & 51 & 54 & 55 & 60 & \ldots \end{array}$$
 
 Você é informado de que $g(1.000) = 2.524$ e $g(1.000.000) = 2.624.152$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md
index d33d9d9942b..68d8540e99a 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md
@@ -24,7 +24,8 @@ Considere $T(N) = \sum_{R = 3}^N \sum_{r=1}^{\left\lfloor \frac{R - 1}{2} \right
 
 Você é informado de que:
 
-$$\begin{align} C(3, 1) = & \\{(3, 0), (-1, 2), (-1,0), (-1,-2)\\} \\\\ C(2500, 1000) = & \\{(2500, 0), (772, 2376), (772, -2376), (516, 1792), (516, -1792), (500, 0), (68, 504), \\\\ &(68, -504),(-1356, 1088), (-1356, -1088), (-1500, 1000), (-1500, -1000)\\} \end{align}$$
+$$\begin{align}   C(3, 1) = & \\{(3, 0), (-1, 2), (-1,0), (-1,-2)\\} \\\\
+  C(2500, 1000) = & \\{(2500, 0), (772, 2376), (772, -2376), (516, 1792), (516, -1792), (500, 0), (68, 504), \\\\ &(68, -504),(-1356, 1088), (-1356, -1088), (-1500, 1000), (-1500, -1000)\\} \end{align}$$
 
 **Observação:** (-625, 0) não é um elemento de $C(2500, 1000)$, pois $\sin(t)$ não é um número racional para os valores correspondentes de $t$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md
index dae6667b25c..a53dd79ed8c 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md
@@ -14,7 +14,10 @@ Há oito números positivos inferiores a 15 que são coprimos para 15: 1, 2, 4,
 
 As inversas modulares desses números modulo 15 são: 1, 8, 4, 13, 2, 11, 7, 14, porque
 
-$$\begin{align} & 1  \times 1\bmod 15 = 1 \\\\ & 2  \times 8  = 16\bmod 15 = 1 \\\\ & 4  \times 4  = 16\bmod 15 = 1 \\\\ & 7  \times 13 = 91\bmod 15 = 1 \\\\ & 11 \times 11 = 121\bmod 15 = 1 \\\\ & 14 \times 14 = 196\bmod 15 = 1 \end{align}$$
+$$\begin{align}   & 1  \times 1\bmod 15 = 1 \\\\
+  & 2  \times 8  = 16\bmod 15 = 1 \\\\   & 4  \times 4  = 16\bmod 15 = 1 \\\\
+  & 7  \times 13 = 91\bmod 15 = 1 \\\\   & 11 \times 11 = 121\bmod 15 = 1 \\\\
+  & 14 \times 14 = 196\bmod 15 = 1 \end{align}$$
 
 Considere $I(n)$ como o maior número positivo $m$ menor que $n - 1$, tal que a inversa modular de $m$ modulo $n$ é igual ao próprio $m$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md
index a9fea6b3d09..200effe6b46 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md
@@ -12,7 +12,9 @@ Considere $f(n)$ como o maior número inteiro positivo $x$ inferior a ${10}^9$,
 
 Por exemplo:
 
-$$\begin{align} & f(4) = 411.728.896 (4^{411.728.896} = ...490\underline{411728896}) \\\\ & f(10) = 0 \\\\ & f(157) = 743.757 (157^{743.757} = ...567\underline{000743757}) \\\\ & Σf(n), 2 ≤ n ≤ 103 = 442.530.011.399 \end{align}$$
+$$\begin{align}   & f(4) = 411.728.896 (4^{411.728.896} = ...490\underline{411728896}) \\\\
+  & f(10) = 0 \\\\   & f(157) = 743.757 (157^{743.757} = ...567\underline{000743757}) \\\\
+  & Σf(n), 2 ≤ n ≤ 103 = 442.530.011.399 \end{align}$$
 
 Encontre $\sum f(n)$, $2 ≤ n ≤ {10}^6$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md
index 36638c38472..811da1817c5 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md
@@ -10,7 +10,8 @@ dashedName: problem-456-triangles-containing-the-origin-ii
 
 Definição:
 
-$$\start{align} & x_n = ({1248}^n\bmod 32323) - 16161 \\\\ & y_n = ({8421}^n\bmod 30103) - 15051 \\\\ & P_n = \\{(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\\} \end{align}$$
+$$\start{align}   & x_n = ({1248}^n\bmod 32323) - 16161 \\\\
+  & y_n = ({8421}^n\bmod 30103) - 15051 \\\\ & P_n = \\{(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\\} \end{align}$$
 
 Por exemplo, $$P_8 = \\{(-14913, -6630), (-10161, 5625), (5226, 11896), (8340, -10778), (15852, -5203), (-15165, 11295), (-1427, -14495), (12407, 1060)\\}$$
 
@@ -18,7 +19,8 @@ Considere $C(n)$ o número de triângulos cujos vértices estão em $P_n$ e que
 
 Exemplos:
 
-$$\start{align} & C(8) = 20 \\\\ & C(600) = 8.950.634 \\\\ & C(40.000) = 266.610.948.988 \end{align}$$
+$$\start{align}   & C(8) = 20 \\\\
+  & C(600) = 8.950.634 \\\\ & C(40.000) = 266.610.948.988 \end{align}$$
 
 Encontre $C(2.000.000)$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md
index 34081959fae..e59494ce320 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md
@@ -10,7 +10,9 @@ dashedName: problem-463-a-weird-recurrence-relation
 
 A função $f$ é definida para todos os números inteiros positivos da seguinte forma:
 
-$$\begin{align} & f(1) = 1 \\\\ & f(3) = 3 \\\\ & f(2n) = f(n) \\\\ & f(4n + 1) = 2f(2n + 1) - f(n) \\\\ & f(4n + 3) = 3f(2n + 1) - 2f(n) \end{align}$$
+$$\begin{align}   & f(1) = 1 \\\\
+  & f(3) = 3 \\\\   & f(2n) = f(n) \\\\
+  & f(4n + 1) = 2f(2n + 1) - f(n) \\\\ & f(4n + 3) = 3f(2n + 1) - 2f(n) \end{align}$$
 
 A função $S(n)$ é definida como $\sum_{i=1}^{n} f(i)$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md
index c06cb3f95aa..bc40a6fe303 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md
@@ -12,13 +12,17 @@ Considere $P(m,n)$ como o número de termos distintos em uma tabela de multiplic
 
 Por exemplo, uma tabela de multiplicação 3×4 fica assim:
 
-$$\begin{array}{c} ×          & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4}  \\\\ \mathbf{1} & 1          & 2          & 3          & 4  \\\\ \mathbf{2} & 2          & 4          & 6          & 8  \\\\ \mathbf{3} & 3          & 6          & 9          & 12 \end{array}$$
+$$\begin{array}{c}   ×          & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4}  \\\\
+  \mathbf{1} & 1          & 2          & 3          & 4  \\\\   \mathbf{2} & 2          & 4          & 6          & 8  \\\\
+  \mathbf{3} & 3          & 6          & 9          & 12 \end{array}$$
 
 Existem 8 termos distintos {1, 2, 3, 4, 6, 8, 9, 12}, portanto $P(3, 4) = 8$.
 
 Você é informado de que:
 
-$$\begin{align} & P(64, 64) = 1.263, \\\\ & P(12, 345) = 1.998, \text{ e} \\\\ & P(32, {10}^{15}) = 13.826.382.602.124.302. \\\\ \end{align}$$
+$$\begin{align}   & P(64, 64) = 1.263, \\\\
+  & P(12, 345) = 1.998, \text{ e} \\\\   & P(32, {10}^{15}) = 13.826.382.602.124.302. \\\\
+\end{align}$$
 
 Encontre $P(64, {10}^{16})$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-467-superinteger.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-467-superinteger.md
index 9e91ca31f8f..9f0930e8205 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-467-superinteger.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-467-superinteger.md
@@ -14,15 +14,18 @@ Por exemplo, 2718281828 é um superinteiro de 18828, enquanto 314159 não é um
 
 Considere $p(n)$ como o número primo $n$ e $c(n)$ como o $n$º número composto. Por exemplo, $p(1) = 2$, $p(10) = 29$, $c(1) = 4$ e $c(10) = 18$.
 
-$$\begin{align} & \\{p(i) : i ≥ 1\\} = \\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \ldots \\} \\\\ & \\{c(i) : i ≥ 1\\} = \\{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, \ldots \\} \end{align}$$
+$$\begin{align}   & \\{p(i) : i ≥ 1\\} = \\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \ldots \\} \\\\
+  & \\{c(i) : i ≥ 1\\} = \\{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, \ldots \\} \end{align}$$
 
 Considere $P^D$ como a sequência de raízes dos algarismos de $\\{p(i)\\}$ ($C^D$ é definido da mesma forma para $\\{c(i)\\}$):
 
-$$\begin{align} & P^D = \\{2, 3, 5, 7, 2, 4, 8, 1, 5, 2, \ldots \\} \\\\ & C^D = \\{4, 6, 8, 9, 1, 3, 5, 6, 7, 9, \ldots \\} \end{align}$$
+$$\begin{align}   & P^D = \\{2, 3, 5, 7, 2, 4, 8, 1, 5, 2, \ldots \\} \\\\
+  & C^D = \\{4, 6, 8, 9, 1, 3, 5, 6, 7, 9, \ldots \\} \end{align}$$
 
 Considere $P_n$ como o número inteiro formado concatenando os primeiros $n$ elementos de $P^D$ ($C_n$ é definido de forma semelhante para $C^D$).
 
-$$\begin{align} & P_{10} = 2.357.248.152 \\\\ & C_{10} = 4.689.135.679 \end{align}$$
+$$\begin{align}   & P_{10} = 2.357.248.152 \\\\
+  & C_{10} = 4.689.135.679 \end{align}$$
 
 Considere $f(n)$ como o menor número inteiro positivo que seja um superinteiro comum de $P_n$ e $C_n$. Por exemplo, $f(10) = 2.357.246.891.352.679$ e $f(100)\bmod 1.000.000.007 = 771.661.825$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md
index e53fba9527f..f9b10cd8ecc 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md
@@ -14,13 +14,15 @@ Considere $SB(n)$ como o maior divisor harmonizado de B de $n$.
 
 Exemplos:
 
-$$\begin{align} & S_1(10) = 1 \\\\ & S_4(2.100) = 12 \\\\ & S_{17}(2.496.144) = 5.712 \end{align}$$
+$$\begin{align}   & S_1(10) = 1 \\\\
+  & S_4(2.100) = 12 \\\\ & S_{17}(2.496.144) = 5.712 \end{align}$$
 
 Defina $F(n) = \displaystyle\sum_{B = 1}^n \sum_{r = 0}^n S_B(\displaystyle\binom{n}{r})$. Aqui, $\displaystyle\binom{n}{r}$ denota o coeficiente binomial.
 
 Exemplos:
 
-$$\begin{align} & F(11) = 3132 \\\\ & F(1.111)\bmod 1.000.000.993 = 706.036.312 \\\\ & F(111.111)\bmod 1.000.000.993 = 22.156.169 \end{align}$$
+$$\begin{align}   & F(11) = 3132 \\\\
+  & F(1.111)\bmod 1.000.000.993 = 706.036.312 \\\\ & F(111.111)\bmod 1.000.000.993 = 22.156.169 \end{align}$$
 
 Encontre $F(11.111.111)\bmod 1.000.000.993$.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md
index fcbfd7496f1..46ad750165e 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md
@@ -16,7 +16,21 @@ Suponha que aquelas com 15 letras ou menos estão listadas em ordem alfabética
 
 A lista incluiria:
 
-$$\begin{align} & 1: \text{a} \\\\ & 2: \text{aa} \\\\ & 3: \text{aaa} \\\\ & 4: \text{aaaa} \\\\ & 5: \text{aaaaa} \\\\ & 6: \text{aaaaaa} \\\\ & 7: \text{aaaaaac} \\\\ & 8: \text{aaaaaacd} \\\\ & 9: \text{aaaaaacde} \\\\ & 10: \text{aaaaaacdee} \\\\ & 11: \text{aaaaaacdeee} \\\\ & 12: \text{aaaaaacdeeee} \\\\ & 13: \text{aaaaaacdeeeee} \\\\ & 14: \text{aaaaaacdeeeeee} \\\\ & 15: \text{aaaaaacdeeeeeef} \\\\ & 16: \text{aaaaaacdeeeeeeg} \\\\ & 17: \text{aaaaaacdeeeeeeh} \\\\ & \ldots \\\\ & 28: \text{aaaaaacdeeeeeey} \\\\ & 29: \text{aaaaaacdeeeeef} \\\\ & 30: \text{aaaaaacdeeeeefe} \\\\ & \ldots \\\\ & 115246685191495242: \text{euleoywuttttsss} \\\\ & 115246685191495243: \text{euler} \\\\ & 115246685191495244: \text{eulera} \\\\ & ... \\\\ & 525069350231428029: \text{ywuuttttssssrrr} \\\\ \end{align}$$
+$$\begin{align}   & 1: \text{a} \\\\
+  & 2: \text{aa} \\\\   & 3: \text{aaa} \\\\
+  & 4: \text{aaaa} \\\\   & 5: \text{aaaaa} \\\\
+  & 6: \text{aaaaaa} \\\\   & 7: \text{aaaaaac} \\\\
+  & 8: \text{aaaaaacd} \\\\   & 9: \text{aaaaaacde} \\\\
+  & 10: \text{aaaaaacdee} \\\\   & 11: \text{aaaaaacdeee} \\\\
+  & 12: \text{aaaaaacdeeee} \\\\   & 13: \text{aaaaaacdeeeee} \\\\
+  & 14: \text{aaaaaacdeeeeee} \\\\   & 15: \text{aaaaaacdeeeeeef} \\\\
+  & 16: \text{aaaaaacdeeeeeeg} \\\\   & 17: \text{aaaaaacdeeeeeeh} \\\\
+  & \ldots \\\\   & 28: \text{aaaaaacdeeeeeey} \\\\
+  & 29: \text{aaaaaacdeeeeef} \\\\   & 30: \text{aaaaaacdeeeeefe} \\\\
+  & \ldots \\\\   & 115246685191495242: \text{euleoywuttttsss} \\\\
+  & 115246685191495243: \text{euler} \\\\   & 115246685191495244: \text{eulera} \\\\
+  & ... \\\\   & 525069350231428029: \text{ywuuttttssssrrr} \\\\
+\end{align}$$
 
 Defina $P(w)$ como a posição da palavra $w$.
 
@@ -26,7 +40,9 @@ Podemos ver que $P(w)$ e $W(p)$ são inversos: $P(W(p)) = p$ e $W(P(w)) = w$.
 
 Exemplos:
 
-$$\begin{align} & W(10) = \text{ aaaaaacdee} \\\\ & P(\text{aaaaaacdee}) = 10 \\\\ & W(115246685191495243) = \text{ euler} \\\\ & P(\text{euler}) = 115246685191495243 \\\\ \end{align}$$
+$$\begin{align}   & W(10) = \text{ aaaaaacdee} \\\\
+  & P(\text{aaaaaacdee}) = 10 \\\\   & W(115246685191495243) = \text{ euler} \\\\
+  & P(\text{euler}) = 115246685191495243 \\\\ \end{align}$$
 
 Encontre $$W(P(\text{legionary}) + P(\text{calorimeters}) - P(\text{annihilate}) + P(\text{orchestrated}) - P(\text{fluttering})).$$
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md
index 5bf0a28020f..5c6c68f0de4 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md
@@ -14,11 +14,15 @@ $$1! + 4! + 5! = 1 + 24 + 120 = 145$$
 
 Talvez 169 seja menos conhecido. Esse número produz a maior cadeia de números que remonta a 169. Acontece que existem apenas três desses laços:
 
-$$\begin{align} &169 → 363601 → 1454 → 169\\\\ &871 → 45361 → 871\\\\ &872 → 45362 → 872\\\\ \end{align}$$
+$$\begin{align} &169 → 363601 → 1454 → 169\\\\
+&871 → 45361 → 871\\\\ &872 → 45362 → 872\\\\
+\end{align}$$
 
 Não é difícil provar que TODOS os números com que você iniciar ficarão presos em um ciclo. Por exemplo:
 
-$$\begin{align} &69 → 363600 → 1454 → 169 → 363601\\ (→ 1454)\\\\ &78 → 45360 → 871 → 45361\\ (→ 871)\\\\ &540 → 145\\ (→ 145)\\\\ \end{align}$$
+$$\begin{align} &69 → 363600 → 1454 → 169 → 363601\\ (→ 1454)\\\\
+&78 → 45360 → 871 → 45361\\ (→ 871)\\\\ &540 → 145\\ (→ 145)\\\\
+\end{align}$$
 
 O número 69 produz uma cadeia de cinco termos sem repetição. A cadeia de maior número sem repetição, iniciando com um número abaixo de um milhão, é de sessenta termos.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md
index de82c18d3c3..c0eb6e85d98 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md
@@ -10,7 +10,9 @@ dashedName: problem-81-path-sum-two-ways
 
 Na matriz de 5 por 5 abaixo, a soma do caminho mínimo do canto superior esquerdo até o canto inferior direito, **movendo-se somente para a direita e para baixo**, é indicado em vermelho e em negrito e é igual a `2427`.
 
-  $$\begin{pmatrix} \color{red}{131} & 673 & 234 & 103 & 18\\\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\\\ 630 & 803 & \color{red}{746} & \color{red}{422} & 111\\\\ 537 & 699 & 497 & \color{red}{121} & 956\\\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$
+  $$\begin{pmatrix}   \color{red}{131} & 673 & 234 & 103 & 18\\\\
+  \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\\\   630 & 803 & \color{red}{746} & \color{red}{422} & 111\\\\
+  537 & 699 & 497 & \color{red}{121} & 956\\\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$
 
 Encontre a soma do caminho mínimo, do canto superior esquerdo para o canto inferior direito, movendo-se apenas para a direita e para baixo, na `matrix`, um array bidimensional que representa uma matriz. O tamanho máximo da matriz utilizado nos testes será de 80 por 80.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md
index e12500f188f..28e395d6e7c 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md
@@ -12,7 +12,9 @@ dashedName: problem-82-path-sum-three-ways
 
 A soma do caminho mínimo da matriz de 5 por 5 abaixo, iniciando em qualquer célula na coluna da esquerda e terminando em qualquer célula na coluna da direita, e apenas se movendo para cima, para baixo e para a direita, é indicado em vermelho e em negrito. A soma é igual a `994`.
 
-  $$\begin{pmatrix} 131 & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\\\ 630 & 803 & 746 & 422 & 111\\\\ 537 & 699 & 497 & 121 & 956\\\\ 805 & 732 & 524 & 37 & 331 \end{pmatrix}$$
+  $$\begin{pmatrix}   131 & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\\\
+  \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\\\   630 & 803 & 746 & 422 & 111\\\\
+  537 & 699 & 497 & 121 & 956\\\\ 805 & 732 & 524 & 37 & 331 \end{pmatrix}$$
 
 Encontre a soma do caminho mínimo, da coluna da esquerda para a coluna da direita, na `matrix`, um array bidimensional que representa uma matriz. O tamanho máximo da matriz utilizado nos testes será de 80 por 80.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md
index 1e0aefd6c04..14c8d7efd5d 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md
@@ -12,7 +12,9 @@ dashedName: problem-83-path-sum-four-ways
 
 Na matriz de 5 por 5 abaixo, a soma do caminho mínimo do canto superior esquerdo até o canto inferior direito, movendo-se para a esquerda, para a direita, para cima e para baixo, é indicado em vermelho e em negrito e é igual a `2297`.
 
-  $$\begin{pmatrix} \color{red}{131} & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & \color{red}{150}\\\\ 630 & 803 & 746 & \color{red}{422} & \color{red}{111}\\\\ 537 & 699 & 497 & \color{red}{121} & 956\\\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$
+  $$\begin{pmatrix}   \color{red}{131} & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\\\
+  \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & \color{red}{150}\\\\   630 & 803 & 746 & \color{red}{422} & \color{red}{111}\\\\
+  537 & 699 & 497 & \color{red}{121} & 956\\\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$
 
 Encontre a soma do caminho mínimo, do canto superior esquerdo para o canto inferior direito, movendo-se para a esquerda, para a direita, para cima e para baixo, na `matrix`, um array bidimensional que representa uma matriz. O tamanho máximo da matriz utilizado nos testes será de 80 por 80.
 
diff --git a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md
index 24ce11a7f0b..af82673a8eb 100644
--- a/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md
+++ b/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md
@@ -12,7 +12,8 @@ Uma cadeia de números é criada adicionando continuamente o quadrado dos algari
 
 Por exemplo:
 
-$$\begin{align} & 44 → 32 → 13 → 10 → \boldsymbol{1} → \boldsymbol{1}\\\\ & 85 → \boldsymbol{89} → 145 → 42 → 20 → 4 → 16 → 37 → 58 → \boldsymbol{89}\\\\ \end{align}$$
+$$\begin{align}   & 44 → 32 → 13 → 10 → \boldsymbol{1} → \boldsymbol{1}\\\\
+  & 85 → \boldsymbol{89} → 145 → 42 → 20 → 4 → 16 → 37 → 58 → \boldsymbol{89}\\\\ \end{align}$$
 
 Portanto, qualquer corrente que chegue a 1 ou 89 ficará presa numa repetição infinita. O que é mais incrível é que TODO número inicial eventualmente chegará a 1 ou 89.