# 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.

def dig_pow(n, p):
    digits = get_digits(n)

    total = 0
    for idx, itm in enumerate(digits):
        total += itm**(p+idx)

    k = total/n
    if k == round(k):
        return k

    return -1

def get_digits(n):
    return [int(ch) for ch in str(n)]

print('dig_pow(46288, 3) should be 51. It is = {}'.format(dig_pow(46288, 3)))
print('dig_pow(92, 1) should be -1. It is = {}'.format(dig_pow(92, 1)))
print('dig_pow(89, 1) should be 1. It is = {}'.format(dig_pow(89, 1)))
