A Number Between Two Others

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:

$$O(t)$$