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.

Steps.

- Lets Convert n which is a number to an Array and to to that we have to convert to a string first.
- Let create a variable total which will hold the sum of all numbers in the array.
- lets create our loop which will iterate over the array we created in step 1
- inside the array we are going to raise each number to the power of p then add it to the variable total we created in step 2
- we will increase our p by 1 because the power goes in an increasing order.
- now we need to find a number which when multiplied with n will give us total. the best way to do that is to divide total by n.
- now lets do our comparison. if the dividend is an integer we will return it, else we return -1 which means no number was found.

Solution.

```
function digPow(n, p){
let nArr = n.toString().split("")
let total = 0
for (let index = 0; index < nArr.length; index++) {
const element = nArr[index];
total += Math.pow(parseInt(element), p)
p++
}
let compare = total / n
if(Number.isInteger(compare)){
return compare
}else{
return -1
}
}
console.log(digPow(46288, 3));
```

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