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Alfredo Salzillo
Alfredo Salzillo

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Can you hack this? #2

#challenge #fun #javascript

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
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Top comments (10)

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pybash profile image
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
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ironcladdev profile image
IroncladDev •

Instead of doing

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

return ans == true;
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omicron666 profile image
Omicron • 
isEven=n=>!!(+n+1<<31)
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vicainelli profile image
Vinicius Cainelli • • Edited

Without convert to Number

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

Can anyone solve it without converting into a number?

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ironcladdev profile image
IroncladDev •

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

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alfredosalzillo profile image
Alfredo Salzillo •

A Little help, you can use a regexp

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idodav profile image
Ido David •

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

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spino233 profile image
Angelo Caci •

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