2016 AMC 10A Problem 25

Step 1: factor. $72 = 2^3 \cdot 3^2$, $600 = 2^3 \cdot 3 \cdot 5^2$ and $900 = 2^2 \cdot 3^2 \cdot 5^2$. Every prime in $x$, $y$ or $z$ must show up in these lcms, so only $2$, $3$ and $5$ can appear.

Step 2: the lcm rule. For each prime, the lcm uses the larger of the two exponents. For example $\text{lcm}(12, 18) = \text{lcm}(2^2 \cdot 3,\ 2 \cdot 3^2) = 2^2 \cdot 3^2 = 36$. So the conditions split up prime by prime, and we can count choices for each prime separately and multiply.

For each prime, call the exponents in $x, y, z$ by the letters $a, b, c$.

Prime 2. The conditions are: larger of $a, b$ is $3$; larger of $a, c$ is $3$; larger of $b, c$ is $2$.
- The last condition says $b$ and $c$ are both at most $2$. So the $3$ in the first two conditions must come from $a$: $a = 3$.
- Now $b, c$ are each $0$, $1$ or $2$, and at least one of them is $2$. Of the $3 \cdot 3 = 9$ pairs, the ones with no $2$ number $2 \cdot 2 = 4$, so $9 - 4 = 5$ pairs work.
- $5$ choices.

Prime 3. Larger of $a, b$ is $2$; larger of $a, c$ is $1$; larger of $b, c$ is $2$.
- The middle condition says $a$ and $c$ are at most $1$. So the $2$ in the first condition must come from $b$: $b = 2$. The third condition then holds automatically.
- Now $a, c$ are each $0$ or $1$ with at least one equal to $1$: the pairs $(1,0)$, $(0,1)$, $(1,1)$.
- $3$ choices.

Prime 5. Larger of $a, b$ is $0$, so $a = b = 0$. Then larger of $a, c$ is $2$ forces $c = 2$, and the last condition checks out.
- $1$ choice.

Step 3: multiply. The choices for different primes do not affect each other, so there are $5 \cdot 3 \cdot 1 = 15$ triples.

The answer is $\boxed{\textbf{(A)}\ 15}$.

Lcm and gcd conditions are really statements about exponents, one prime at a time, and different primes never interact, so the answer is a product of small counts. Within one prime, three "larger of two" conditions usually force one exponent right away and leave a simple count for the other two. For "the larger of two numbers from $0$ to $t$ is exactly $t$", count all pairs and subtract the ones with no $t$.