Playing with digits

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 as 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

My solution:

function digPow(n, p){
  let array = [];
  let m = n;
  
  function reducer(acc,curr, index) {
    acc = acc + Math.pow(curr, index+p);
    return acc;
  }
  
  while(true) {
    let unit = n % 10;
    array.push(unit);
    
    n = (n-unit) / 10;
    if(n == 0) {
      break;
    }
  }
  
  array = array.reverse();
  const testNumber = array.reduce(reducer, 0);
  const k = testNumber / m;
  
  if(Number.isInteger(k) == true) {
    return k;
  }
  else {
    return -1; 
  }
  
}

Best solutions:

function digPow(n, p) {
  var x = String(n).split("").map(a=>parseInt(a)).reduce((s, d, i) => s + Math.pow(d, p + i), 0)
  return x % n ? -1 : x / n
}

• String(number).split("").map(a⇒parseInt(a))

  각 자릿수를 요소로 하는 배열을 만들기 위하여서 먼저 숫자를 문자열로 바꾼뒤에 각 문자를 요소로 하는 배열을 만든다. 그 뒤에 다시 각 요소들을 숫자로 바꾼다.
  

• condition ? exprIfTrue : exprIfFalse

  shorthanded version of if문. 해당 코드에서는 나머지가 0이면 조건이 false가 되어서 exprIfFalse를 실행하고, 나머지 경우에는 true가 되어서 exprIfTrue를 실행한다.