freeCodeCamp/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-multiplication.md

id, title, challengeType, videoUrl, forumTopicId, dashedName

id	title	challengeType	videoUrl	forumTopicId	dashedName
56533eb9ac21ba0edf2244b1	Compound Assignment With Augmented Multiplication	1	https://scrimba.com/c/c83vrfa	16662	compound-assignment-with-augmented-multiplication

--description--

The *= operator multiplies a variable by a number.

myVar = myVar * 5;

will multiply myVar by 5. This can be rewritten as:

myVar *= 5;

--instructions--

Convert the assignments for a, b, and c to use the *= operator.

--hints--

a should equal 25.

assert(a === 25);

b should equal 36.

assert(b === 36);

c should equal 46.

assert(c === 46);

You should use the *= operator for each variable.

assert(__helpers.removeJSComments(code).match(/\*=/g).length === 3);

You should not modify the code above the specified comment.

assert(
  /let a = 5;/.test(__helpers.removeJSComments(code)) &&
    /let b = 12;/.test(__helpers.removeJSComments(code)) &&
    /let c = 4\.6;/.test(__helpers.removeJSComments(code))
);

--seed--

--after-user-code--

(function(a,b,c){ return "a = " + a + ", b = " + b + ", c = " + c; })(a,b,c);

--seed-contents--

let a = 5;
let b = 12;
let c = 4.6;

// Only change code below this line
a = a * 5;
b = 3 * b;
c = c * 10;

--solutions--

let a = 5;
let b = 12;
let c = 4.6;

a *= 5;
b *= 3;
c *= 10;