freeCodeCamp/curriculum/challenges/english/99-dev-playground/daily-coding-challenges-python/python-challenge-10.md

id, title, challengeType, dashedName

id	title	challengeType	dashedName
681cb1b3dab50c87ddb2e524	Python Challenge 10: 3 Strikes	29	python-challenge-10

--description--

Given an integer between 1 and 10,000, return a count of how many numbers from 1 up to that integer whose square contains at least one digit 3.

--hints--

squares_with_three(1) should return 0.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(squares_with_three(1), 0)`)
}})

squares_with_three(10) should return 1.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(squares_with_three(10), 1)`)
}})

squares_with_three(100) should return 19.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(squares_with_three(100), 19)`)
}})

squares_with_three(1000) should return 326.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(squares_with_three(1000), 326)`)
}})

squares_with_three(10000) should return 4531.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(squares_with_three(10000), 4531)`)
}})

--seed--

--seed-contents--

def squares_with_three(n):

    return n

--solutions--

def squares_with_three(n):
    count = 0
    for i in range(1, n + 1):
        square = i * i
        if '3' in str(square):
            count += 1
    return count