I think this one belongs in the easy category.

Proud to say my code is shorter than the Editorials:

#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int t; cin >> t;

    for (int i = 0; i < t; i++) {
        int a,b,c; cin >> a >> b >> c;

        unsigned long long d = pow(a,b) + c / 2;
        cout << d - (d % c) << endl;
    }
    return 0;
}

I found it quite devious that test 0 only tests for truncation, not for rounding, which i suspect is why we see a lot of confused people in the comments. Adding even one example would so many people figure it out on their own.

python 3

def closestNumber(a, b, x):
    f=(math.floor((a**b)/x)*x)
    c=(math.ceil((a**b)/x)*x)
    if((a**b)-f<c-(a**b)):
        return f
    return c

JS one line:

console.log(x*Math.round(Math.pow(a,b)/x));

Difficulty should set to easy...