2225AUnreviewedunratedGPT-5.5 (high)

A Number Between Two Others

Progressive hints first, then the full explanation and implementation when you're ready to cash out.

greedy math

Original problem

Review status

AI-generated and still unreviewed. Double-check the details before internalizing them.

Hints

Progressive nudges

Open only as much as you need to keep the solve alive.

Because both $y$ and the desired $z$ are multiples of $x$ , try factoring out $x$ completely. The actual size of $x$ is mostly a distraction.

Write $y = kx$ and $z = mx$ . Then $x < z < y$ becomes $1 < m < k$ .

The divisibility condition also simplifies: $z \mid y$ is equivalent to $m \mid k$ . So we need some $m$ with $1 < m < k$ such that $m \nmid k$ .

For $k > 2$ , look at the largest possible choice: $m = k - 1$ . Does $k - 1$ divide $k$ ?

If $y = 2x$ , there is no multiple of $x$ strictly between them, so answer is NO. Otherwise $z = y - x$ works, because $y \bmod (y-x) = x \ne 0$ .

Key observation

Since $y$ is divisible by $x$ , write

y = kx

where $k$ is an integer and $k > 1$ .

Any valid $z$ must also be divisible by $x$ , so it has the form

z = mx

for some integer $m$ .

Now translate the conditions:

$x < z < y$ becomes

1 < m < k

$y$ is divisible by $z$ means

mx \mid kx \iff m \mid k

So the problem is now hilariously small:

Does there exist an integer $m$ such that $1 < m < k$ and $m \nmid k$ ?

The entire trick

If $k = 2$ , then there is no integer $m$ with

1 < m < 2

So the answer is definitely NO.

If $k > 2$ , choose

m = k - 1

Then $m$ is valid because

1 < k - 1 < k

Also, $k - 1$ does not divide $k$ , since

k = 1 \cdot (k - 1) + 1

so the remainder is $1$ .

Returning to the original numbers, this choice corresponds to

z = (k-1)x = y - x

For $k > 2$ , this satisfies $x < z < y$ , and

y \bmod (y-x) = x \ne 0

So it works. Clean. Almost suspiciously clean, but that's the whole problem.

Final criterion

Since $k = \frac{y}{x}$ :

\text{answer is YES} \iff \frac{y}{x} > 2

Otherwise, answer is NO.

Complexity

Each test case is solved with one division and one comparison.

O(1) \text{ per test case}

Total complexity:

O(t)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() {
    setIO();

    int t;
    cin >> t;

    while (t--) {
        ll x, y;
        cin >> x >> y;

        ll k = y / x;
        cout << (k > 2 ? "YES" : "NO") << '\n';
    }

    return 0;
}

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() {
    setIO();

    int t;
    cin >> t;

    while (t--) {
        ll x, y;
        cin >> x >> y;

        ll k = y / x;
        cout << (k > 2 ? "YES" : "NO") << '\n';
    }

    return 0;
}

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