Hey everyone,
While solving Problem A (A Number Between Two Others), I noticed a lot of solutions get caught up thinking about prime factors or trying to construct specific test values for $$$z$$$.
Turns out, the entire problem collapses into a single inequality.
The Shortcut
Let $$$y = k \cdot x$$$. Since $$$y$$$ is divisible by $$$x$$$, $$$k \ge 2$$$.
We need a multiple of $$$x$$$, say $$$z = m \cdot x$$$, that lies between $$$x$$$ and $$$y$$$ and does not divide $$$y$$$.
This just means finding an integer $$$m$$$ between $$$1$$$ and $$$k$$$ where $$$k$$$ is not divisible by $$$m$$$:
If $$$k = 2$$$: No integer exists strictly between $$$1$$$ and $$$2$$$ $$$\to$$$ NO.
If $$$k \ge 3$$$: Always possible $$$\to$$$ YES.
So the answer is YES if and only if $$$k \gt 2$$$, which simply means: b - a > a (or b > 2 * a).
Code
```cpp
#include <bits/stdc++.h>
using namespace std;
#define ll long long
void solve(){
ll a,b;
cin>>a>>b;
cout<<(b-a>a ? "YES\n" : "NO\n");
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while (t--) solve();
return 0;
}








k >= 3 works because We can always find a number z = (k-1)*x which is never divisible by y = k*x because k % (k-1) != 0 for k >= 3.
Yaaa