AMC 12 Mock Exam 5 Problem 18

Write $600=2^3\cdot3\cdot5^2$. For a prime $p$ with exponent $e$ in $600$, the exponents of $p$ in $a,b,c$ form a triple in $\{0,\dots,e\}^3$ whose maximum is $e$ and whose minimum is $0$. By inclusion–exclusion the number of such triples is $$(e+1)^3-2e^3+(e-1)^3,$$ (all triples, minus those missing $e$, minus those missing $0$, plus those missing both). This is $64-54+8=18$ for $e=3$, $8-2+0=6$ for $e=1$, and $27-16+1=12$ for $e=2$. The primes are independent, so the count is $18\cdot6\cdot12=\boxed{1296}$.