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Write an if....else statement that:

-prints "even" if the number is an even number
-prints "odd" if the number is an odd number

(use the % operator to determine if the number is odd or even)

What I have tried:

/*
* Programming Quiz: Even or Odd (3-2)
*
* Write an if...else statement that prints `even` if the
* number is even and prints `odd` if the number is odd.
*
* Note - make sure to print only the string "even" or the string "odd"
*/

// change the value of `number` to test your if...else statement

var number = 2;

if (number % 2 = 0 ); {console.log("even");} else {console.log("odd");

}
Posted
Updated 8-Jan-20 8:01am
Comments
0x01AA 8-Jan-20 12:38pm
   
And what is the problem?
Btw. the first semicolon and the equality operator are wrong. It should be
if (number % 2 == 0 ) {....
instead of
if (number % 2 = 0 ); {....

1 solution

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Solution 1

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Comments
Member 14710674 8-Jan-20 13:06pm
   
It wasn't for a grade. I'm simply teaching myself and I was stuck in the equation part. I already posted how far I had gotten.

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