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

id	title	challengeType	videoUrl
56533eb9ac21ba0edf2244b1	Compound Assignment With Augmented Multiplication	1	https://scrimba.com/c/c83vrfa

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.

Tests

tests:
  - text: <code>a</code> should equal <code>25</code>
    testString: assert(a === 25);
  - text: <code>b</code> should equal <code>36</code>
    testString: assert(b === 36);
  - text: <code>c</code> should equal <code>46</code>
    testString: assert(c === 46);
  - text: You should use the <code>*=</code> operator for each variable
    testString: assert(code.match(/\*=/g).length === 3);
  - text: Do not modify the code above the line
    testString: assert(/var a = 5;/.test(code) && /var b = 12;/.test(code) && /var c = 4\.6;/.test(code));

Challenge Seed

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

// Only modify code below this line

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

After Test

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

Solution

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

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