* Editorialist:* akshayakgec

# DIFFICULTY:

EASY-MEDIUM

# PREREQUISITES:

Basic modulo and number theory

# PROBLEM:

You are given two numbers XX & YY (XX != YY) your task is to determine **count** of such positive numbers KK that the remainders of X and Y when divided by K are the same. More formally you need to determine **count of such K** (K > 0) that **X % K == Y % K** .

# EXPLANATION:

If x and y are not equal then our answer will be the total number of factors absolute difference of x and y.

between x and y.

Given:- (x % k) = (y % k)

((x % k)- (y % k)) = 0

((x % k) - (y % k)) % k=0 % k

We know the property of modulo- ((a % k) - (b % k)) % k = (a - b) % k;

(x - y) % k = 0

If (x - y) is a multiple of k then their modulo will be 0

We have to find the total number of factors of the absolute difference of x and y.

## Editorialist's Solution

```
#include <bits/stdc++.h>
#define ll long long
using namespace std;
int main() {
int t; cin >> t;
while(t--) {
ll x,y;
cin >> x >> y;
ll k;
int cnt = 0;
k = abs(x - y);
for(ll i=1; i<=sqrt(k); i++) {
if(k % i == 0){
cnt++;
if(k / i != i)
{
cnt++;
}
}
}
cout << cnt << "\n";
}
return 0;
}
```