Japan Junior Mathematical Olympiad

The prime factorization of $720$ is given by $2^4\times 3^2\times 5^1$. Therefore, none of $a,b,c$ has prime factors other than $2$, $3$, $5$, so we can write $$a=2^{p_1}{3}^{q_1}{5}^{r_1},b=2^{p_2}{3}^{q_2}{5}^{r_2},c=2^{p_3}{3}^{q_3}{5}^{r_3}.$$ where, for each $i$, $p_i,q_i,r_i$ are nonnegative integers. The fact that the least common multiple of $a,b,c$ is $720=2^4\times 3^2\times 5^1$ is equivalent to the validity of all of the following three conditions:

(1) The largest number among $p_1,p_2,p_3$ is $4$.

(2) The largest number among $q_1,q_2,q_3$ is $2$.

(3) The largest number among $r_1,r_2,r_3$ is $1$.

Let us find the number of the triplets $(p_1,p_2,p_3)$ which satisfy the requirement (1). (We distinguish triplets obtained from the same $3$ numbers by changing their order in the arrangement). There are $5^3=125$ triplets $(p_1,p_2,p_3)$ for which every $p_i$ ($i=1,2,3$) satisfies $0\le p_i\le 4$, and $4^3=64$ triplets for which $0\le p_i\le 3$ for each $i$. The number of those triplets $(p_1,p_2,p_3)$ satisfying the condition (1) is given by the difference of these two numbers, and thus it is $125-64=61$.

Similarly, the number of triplets $(q_1,q_2,q_3)$ satisfying the condition (2) is $3^3-2^3=19$ and the number of triplets $(r_1,r_2,r_3)$ satisfying (3) is $2^3-1^3=7$.

Therefore, the number of triplets $(a,b,c)$ satisfying the condition of the problem is $61\times 19\times 7=8113$.