2016 AMC 10A Problem 25

Factor: $72 = 2^3\cdot 3^2$, $600 = 2^3\cdot 3\cdot 5^2$, $900 = 2^2\cdot 3^2\cdot 5^2$. No other primes can appear in $x, y, z$. Since an lcm takes the larger exponent of each prime, the three conditions separate prime by prime; let $x, y, z$ carry exponents $(a, b, c)$ for the prime under consideration.

Prime 2: $\max(a,b) = 3$, $\max(a,c) = 3$, $\max(b,c) = 2$. Neither $b$ nor $c$ can be $3$, so $a = 3$. Then $b, c \in \{0,1,2\}$ with $\max(b,c) = 2$: $3^2 - 2^2 = 5$ pairs.

Prime 3: $\max(a,b) = 2$, $\max(a,c) = 1$, $\max(b,c) = 2$. Since $a, c \le 1$, we need $b = 2$. Then $a, c \in \{0,1\}$ with $\max(a,c) = 1$: $2^2 - 1^2 = 3$ pairs.

Prime 5: $\max(a,b) = 0$ forces $a = b = 0$, and then $\max(a,c) = 2$ forces $c = 2$: $1$ way.

The exponent choices for different primes are independent, so the number of triples is $5\cdot 3\cdot 1 = 15$.

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

Conditions built from gcd and lcm are statements about exponents, one prime at a time, and the primes never interact, so the answer is a product of small counts. Within each prime, an lcm equation $\max(u,v) = t$ means one of $u, v$ equals $t$ and the other is at most $t$; combining three such equations usually forces one exponent outright and leaves a "max equals $t$" count of $(t+1)^2 - t^2 = 2t+1$ for the other two.

• Counting exponent pairs with max equal to the target as (target + 1)^2 pairs instead of 2(target) + 1, or forgetting that a forced exponent (like 3 for the prime 2 in x) removes freedom.
• Mixing up which variable is forced for each prime (for the prime 3 it is $y$, not $x$), which changes the remaining count.