Playing with digits

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, 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

Good Solution1:

public class DigPow {
  
  public static long digPow(int n, int p) {
    String intString = String.valueOf(n);
    long sum = 0;
    for (int i = 0; i < intString.length(); ++i, ++p)
      sum += Math.pow(Character.getNumericValue(intString.charAt(i)), p);
    return (sum % n == 0) ? sum / n : -1;
  }
  
}

Good Solution2:

class DigPow {
  
  public static long digPow(int n, int p) {
    long s = 0;
    String nstr = String.valueOf(n);
    for (int i = 0; i < nstr.length(); i++) {
      s += (long)Math.pow((int)(nstr.charAt(i) -'0'), p+i);
    }
    if (s % n == 0) 
      return s / n;
    else return -1;
  }
  
}