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.

```python
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
```

Difficulty: introductory

Starter code:

def dig_pow(n, p):
	

Solutions:
def dig_pow(n, p):
  s = 0
  for i,c in enumerate(str(n)):
     s += pow(int(c),p+i)
  return s/n if s%n==0 else -1


def dig_pow(n, p):
    k, fail = divmod(sum(int(d)**(p + i) for i, d in enumerate(str(n))), n)
    return -1 if fail else k


def dig_pow(n, p):
  t = sum( int(d) ** (p+i) for i, d in enumerate(str(n)) )
  return t//n if t%n==0 else -1

def dig_pow(n, p):
    # your code
    n_list = [int(i) for i in str(n)]
    n_sum = 0
    p_i = p
    for n_i in n_list:
        n_sum = n_sum + n_i**p_i
        p_i = p_i+1
    if n_sum%n == 0:
        return n_sum/n
    else:
        return -1

def dig_pow(n, p):
    sum = 0
    for c in str(n):
        sum += int(c) ** p
        p += 1
    if sum % n == 0:
        return sum / n
    else:
        return -1

def dig_pow(n, p):
    s = sum(int(d)**(p+i) for i,d in enumerate(str(n)))
    return s/n if s%n == 0 else -1

def dig_pow(n, p):
    digit_power = sum(int(x)**pw for pw,x in enumerate(str(n),p))
    if digit_power % n == 0:
        return digit_power / n
    return -1

def dig_pow(n, p):
    k, r = divmod(sum( int(l)**(p+i) for i,l in enumerate(str(n)) ), n)
    return -1 if r else k

def dig_pow(n, p):
    digits = [int(d) for d in str(n)]
    
    sum = 0
    for i, d in enumerate(digits):
        sum += d ** (p+i)
    
    if sum % n == 0:
        return sum / n
    else:
        return -1