HMMT February, 2019

Solution:
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
$$12\binom{n+1}{3} + 8\binom{n+1}{2} + (n+1) = 2(n+1)n(n-1) + 4(n+1)n + (n+1) = (n+1)(2n(n+1)+1)$$
possible quadruples. So if we choose a random quadruple then the probability that it satisfies the condition is $\frac{(n+1)(2n(n+1)+1)}{(n+1)^4} = \frac{2n(n+1)+1}{(n+1)^3}$.

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
$$\frac{41}{125} \cdot \frac{25}{64} \cdot \frac{5}{8} = \frac{41}{512}$$