PlayingWithDigits.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):
    answer = 0
    expo = p
    for num in str(n):
        answer += int(num)**expo
        expo += 1
    
    if answer % n == 0:
        return answer / n
    
    return -1


# for this kata we need to start our exponent at p and add one each time. then we add each individual number to answer after 
# its exponential answer. then if answer % n is 0, we know that n goes evenly into 0 and return our k. else -1