Question

Let $a$ be a factor of $120$. If $\lambda$ be the number of positive integral solutions of $x_1{x}_2{x}_3=a$, then sum of the digits in $\lambda$ is

Answer 1

Let $x_1=2^{a_1}\times 3^{b_1}\times 5^{c_1}$
$x_2=2^{a_2}\times 3^{b_2}\times 5^{c_2}$
$x_3=2^{a_3}\times 3^{b_3}\times 5^{c_3}$
Let $x_4=2^{a_4}\times 3^{b_4}\times 5^{c_4}$ such that
$x_1\times x_2\times x_3\times x_4=120$

Number of positive integral solutions of $x_1{x}_2{x}_3=a$, where $a$ is a factor of $120$ is equivalent to number of positive integral solution of $x_1\times x_2\times x_3\times x_4=120$
$120=2^3\times 3\times 5$
Hence,
$a_1+a_2+a_3+a_4=3$...(1) ($0\le a_i\le 3$)
$b_1+b_2+b_3+b_4=1$....(2) ($0\le b_i\le 1$)
$c_1+c_2+c_3+c_4=1$....(3) ($0\le c_i\le 1$)
We can select $a_i$'s independently of $b_i$'s and $c_i$'s.

Number of solutions of (1) is ${}^6{C}_3$

Number of solutions of (2) is ${}^4{C}_3$

Number of solutions of (3) is ${}^4{C}_3$
Hence, total number of ways of selecting ${=}^6{C}_3{\times }^4{C}_3{\times }^4{C}_3$
Hence, total number of ways of selecting $=20\times 4\times 4$
Total number of ways $=320$
Hence, $\lambda =320$
Sum of digits in $\lambda =5$