---
id: 56105e7b514f539506016a5e
title: Count Backwards With a For Loop
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2R6BHa'
---
## Description
A for loop can also count backwards, so long as we can define the right conditions.
In order to count backwards by twos, we'll need to change our initialization, condition, and final-expression.
We'll start at i = 10 and loop while i > 0. We'll decrement i by 2 each loop with i -= 2.
var ourArray = [];
for (var i=10; i > 0; i-=2) {
ourArray.push(i);
}
ourArray will now contain [10,8,6,4,2].
Let's change our initialization and final-expression so we can count backward by twos by odd numbers.
## Instructions
Push the odd numbers from 9 through 1 to myArray using a for loop.
## Tests
```yml
tests:
- text: You should be using a for loop for this.
testString: assert(code.match(/for\s*\(/g).length > 1, 'You should be using a for loop for this.');
- text: You should be using the array method push.
testString: assert(code.match(/myArray.push/), 'You should be using the array method push.');
- text: myArray should equal [9,7,5,3,1].
testString: assert.deepEqual(myArray, [9,7,5,3,1], 'myArray should equal [9,7,5,3,1].');
```
## Challenge Seed
```js
// Example
var ourArray = [];
for (var i = 10; i > 0; i -= 2) {
ourArray.push(i);
}
// Setup
var myArray = [];
// Only change code below this line.
```
### After Test
```js
if(typeof myArray !== "undefined"){(function(){return myArray;})();}
```
## Solution
```js
var ourArray = [];
for (var i = 10; i > 0; i -= 2) {
ourArray.push(i);
}
var myArray = [];
for (var i = 9; i > 0; i -= 2) {
myArray.push(i);
}
```