Can you hack this? #2

Write an isEven function to check if a number is even without using the modulus operator.

const isEven = (n) => ...

isEven(2) // => true 
isEven('127') // => false
isEven('12abc2') // => false 

Comments

[Omicron]

isEven=n=>!!(+n+1<<31)

[PyBash]

Solved!

const isEven = (n) => {
  q = n/2
  ans = q===parseInt(q)
  if (ans==true) {
    return true
  }
  else {
    return false
  }
}

console.log(isEven(2)) // => true
console.log(isEven(127)) // => false
console.log(isEven('127')) // => false
console.log(isEven('12abc2')) // => false

[IroncladDev]

Instead of doing

if (ans==true) {
    return true
  }
  else {
    return false
  }

Try

return ans == true;

[Vinicius Cainelli]

Without convert to Number

const isEven = (n) => {
  const rgx = new RegExp('^\d*[02468]$')
  return rgx.test(n)
};

[Ido David]

another solution:
isEven = (num) => Number.isInteger(num) && (!(num & 1))

[Alfredo Salzillo]

Can anyone solve it without converting into a number?

[IroncladDev]

I don't know if that's possible. I mean, using the parseInt function could work if someone inputs a string.

[Alfredo Salzillo]

A Little help, you can use a regexp

[Angelo Caci]

const isEven = (n) => !!n.toString().match("^[0-9]*[02468]$");