How to determine if a number is odd in JavaScript

Question

Use the below code:

function isOdd(num) { return num % 2;}
console.log("1 is " + isOdd(1));
console.log("2 is " + isOdd(2));
console.log("3 is " + isOdd(3));
console.log("4 is " + isOdd(4));

1 represents an odd number, while 0 represents an even number.

Use the bitwise AND operator.

function oddOrEven(x) {
  return ( x & 1 ) ? "odd" : "even";
}

function checkNumber(argNumber) {
  document.getElementById("result").innerHTML = "Number " + argNumber + " is " + oddOrEven(argNumber);
}
 
checkNumber(17);

<div id="result" style="font-size:150%;text-shadow: 1px 1px 2px #CE5937;" ></div>

If you don't want a string return value, but rather a boolean one, use this:

var isOdd = function(x) { return x & 1; };
var isEven  = function(x) { return !( x & 1 ); };

You could do something like this:

function isEven(value){
    if (value%2 == 0)
        return true;
    else
        return false;
}

function isEven(x) { return (x%2)==0; }
function isOdd(x) { return !isEven(x); }

Do I have to make an array really large that has a lot of even numbers

No. Use modulus (%). It gives you the remainder of the two numbers you are dividing.

Ex. 2 % 2 = 0 because 2/2 = 1 with 0 remainder.

Ex2. 3 % 2 = 1 because 3/2 = 1 with 1 remainder.

Ex3. -7 % 2 = -1 because -7/2 = -3 with -1 remainder.

This means if you mod any number x by 2, you get either 0 or 1 or -1. 0 would mean it's even. Anything else would mean it's odd.

This can be solved with a small snippet of code:

function isEven(value) {
    return !(value % 2)
}

Hope this helps :)

In ES6:

const isOdd = num => num % 2 == 1;