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Suppose there are three variables N, R and P. The N and R are used to get the ^{N}C_{R} and P is a prime. We have to find whether ^{N}C_{R} is divisible by P. Suppose we have some numbers N = 7, R = 2 and P = 3, then ^{7}C_{2} = 21, this is divisible by 3, so the output will be true.

We know that ^{N}C_{R} = N! / (R! * (N – R)! ). We will use Legendre Formula to largest power of P, which divides any N!, R! and (N – R)! in order to NCR to be divisible by P, the condition is N! > R! + (N - R)!

#include <iostream> using namespace std; int getPower(int n, int p) { int pow = 0; while (n) { n /= p; pow += n; } return pow; } bool isDivisibleByP(int n, int r, int p) { // Find the highest powers of p // that divide n!, r! and (n - r)! int x1 = getPower(n, p); int x2 = getPower(r, p); int x3 = getPower(n - r, p); if (x1 > x2 + x3) return true; return false; } int main() { int n = 7, r = 2, p = 7; if (isDivisibleByP(n, r, p)) cout << "nCr is divisible by P"; else cout << "nCr is not divisible by P"; }

nCr is divisible by P

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