Counting Triples II

Note that $9000=2^33^25^3$ . Since $[a,b,c]=9000$ , there exist nonnegative integers $a_1,b_1,c_1,a_2,b_2,c_2,a_3,b_3$ and $c_3$ such that $a=2^{a_1}3^{b_1}5^{c_1},b=2^{a_2}3^{b_2}5^{c_2}$ and $c=2^{a_3}3^{b_3}5^{c_3}$ . Finding the number of ordered triples $N$ satisfying $[a,b,c]=9000$ is equivalent to finding the number of solutions $(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)$ such that $\max\{a_1,a_2,a_3\}=3,\max\{b_1,b_2,b_3\}=2$ and $\max\{c_1,c_2,c_3\}=3$ . We then deduce that at least one of $a_1,a_2,a_3$ is $3$ , so the number of solutions of $\max\{a_1,a_2,a_3\}=3$ is $4^3-3^3=37$ . Similarly, the number of solutions of $\max\{b_1,b_2,b_3\}=2$ is $3^3-2^3=19$ and the number of solutions of $\max\{c_1,c_2,c_3\}=3$ is $4^3-3^3=37$ . Thus, the number of ordered triples $N$ is $N=37^2\cdot 19 = 26011\equiv 11\pmod {1000}$ .