playingDigits.py

# 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.

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

def dig_pow(n, p):
    num = str(n)
    total = sum([int(num[i]) ** (p + i) for i in range(len(num))])
    return total / n if (total % n) == 0 else -1