playing-with-digits.md

Playing with digits ^{6 Kyu}

^{LINK TO THE KATA - FUNDAMENTALS MATHEMATICS}

Description

Some numbers have funny properties. For example:

89 --> 8¹ + 9² = 89 * 1

695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2

46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51

Given a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p

• we want to find a positive integer k, if it exists, such that the sum of the digits of n taken to the successive powers of p is equal to k * n.

In other words:

Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k

If it is the case we will return k, if not return -1.

Note: n and p will always be given as strictly positive integers.

digPow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
digPow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
digPow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
digPow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51

Solution

const digPow = (n, p) => {
  const numbers = n
    .toString()
    .split('')
    .map(character => Number(character))

  const powSum = numbers.reduce((acc, cur, i) => acc + Math.pow(cur, p + i), 0)

  if (powSum === n * p) return p
  if (powSum % n === 0) return powSum / n
  return -1
}