← Back to challenges

Oddish vs. Evenish

JavaScriptHardnumbersmathvalidationconditions

Instructions

Create a function that determines whether a number is Oddish or Evenish. A number is Oddish if the sum of all of its digits is odd, and a number is Evenish if the sum of all of its digits is even. If a number is Oddish, return "Oddish". Otherwise, return "Evenish".

For example, oddishOrEvenish(121) should return "Evenish", since 1 + 2 + 1 = 4. oddishOrEvenish(41) should return "Oddish", since 4 + 1 = 5.

Examples

oddishOrEvenish(43) ➞ "Oddish"
// 4 + 3 = 7
// 7 % 2 = 1

oddishOrEvenish(373) ➞ "Oddish"
// 3 + 7 + 3 = 13
// 13 % 2 = 1

oddishOrEvenish(4433) ➞ "Evenish"
// 4 + 4 + 3 + 3 = 14
// 14 % 2 = 0

Notes

N/A

javascript
Loading editor…
to run
Walks through the solution with reasoning and edge cases.