HMMT February

For each prime factor, let the greatest power that divides $a, b, c, d$ be $p, q, r, s$. WLOG assume that $p \leq q$ and $r \leq s$, and further WLOG assume that $p \leq r$. Then we need $r=\min (q, s)$. If $q=r$ then we have $p \leq q=r \leq s$, and if $r=s$ then we have $p \leq r=s \leq q$, and in either case the condition reduces to the two 'medians' among $p, q, r, s$ are equal. (It is not difficult to see that this condition is also sufficient.) Now we compute the number of quadruples $(p, q, r, s)$ of integers between 0 and $n$ inclusive that satisfy the above condition. If there are three distinct numbers then there are $\binom{n+1}{3}$ ways to choose the three numbers and $4!/ 2=12$ ways to assign them (it must be a $1-2-1$ split). If there are two distinct numbers then there are $\binom{n+1}{2}$ ways to choose the numbers and $4+4=8$ ways to assign them (it must be a $3-1$ or a 1-3 split). If there is one distinct number then there are $n+1$ ways to assign. Together we have $12\binom{n+1}{3}+8\binom{n+1}{2}+(n+1)=2(n+1) n(n-1)+4(n+1) n+(n+1)=(n+1)(2 n(n+1)+1)$ possible quadruples. So if we choose a random quadruple then the probability that it satisfies the condition is $\frac{(n+1)(2 n(n+1)+1)}{(n+1)^{4}}=\frac{2 n(n+1)+1}{(n+1)^{3}}$. Since $6000=2^{4} \cdot 5^{3} \cdot 3^{1}$ and the power of different primes are independent, we plug in $n=4,3,1$ to get the overall probability to be $$\frac{41}{125} \cdot \frac{25}{64} \cdot \frac{5}{8}=\frac{41}{512}$$