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Given two positive integer N and X. The task is to express N as a sum of powers of X (X0 + X1 +…..+ Xn) such that the number of powers of X should be minimum.

Print the minimum number of power of N used to make the sum equal to N.

If N = 15 and X = 3 then we need 3 powers of ‘3’ as follows −

15 = (3^{2} + 3^{1} + 3^{1})

Use below formula to calculate final result −

1. If x = 1, then answer will be n only (n = 1 + 1 +…. n times)s 2. Any number n can be expressed as, n = x * a + b where 0 −= b −= x-1. Now since b is between 0 to x – 1, then b should be expressed as sum of x0 b times

#include <iostream> using namespace std; int minNumOfPower(int n, int x){ if (x == 1) { return n; } int result = 0; while (n > 0) { result = result + (n % x); n = n / x; } return result; } int main(){ int n = 15; int x = 3; cout << "Minimum number of powers = " << minNumOfPower(15, 3) << endl; return 0; }

When you compile and execute the above program. It generates the following output −

Minimum number of powers = 3

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