playing-with-digits.js

// 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

function digPow(n, p) {
  let arrayFromN = Array.from(String(n), Number);
  let x = arrayFromN.length;
  let newArray = [];
  for (i = p; i < x + p; i++) {
    newArray.push(arrayFromN.shift() ** i);
  }
  const arraySum = newArray.reduce((previousValue, currentValue) => previousValue+currentValue)
  if(arraySum%n === 0){
    return arraySum/n
  }
  else{
    return -1
  }
}