Fun Loops ▶️

Because the abs(3) == 3 is true and that isn't even.

An even number of flips would be true and an odd number of flips would be false which works out.

I was thinking a bitwise & or converting it to a string and testing if the right most character is 0, 2, 4, 6, 8 would be panic mode solutions too.

Lol I think the interview pressure got to you.

I'm not following.
n == abs(n) only tells you if it's positive.

That would be isPositive.

Without using the modulo operator you'd essentially have to reimplement it. Divide the number by 2 and round down. Multiply that by 2 and then subtract it from the original number.

isEven(10) results in 10-10==0 (true) whereas isEven(13) results in 13-12==0 (false).

function isEven(n){
  n = Math.abs(n)
  return (n - (Math.floor(n/2) * 2)) == 0
}