Question

The total number of integral solutions for $(x,y,z)$ such that $xyz=24$ is

• A. $36$
• B. $90$
• C. $120$
• D. none of these

Answer 1

The correct option is C $120$

The number $24$ can be factored as: $24=2^3\times 3$. Hence, you need $x,y,z$ to be of the form $2^a{3}^b$ You have 3 powers of 2 and one power of 3 which you need to distribute between three numbers.

For $2^{\prime }s$, this can be done in ${}^5{\text{C}}_2$=10 ways($a_1+a_2+a_3=3$ =>${}^{n+r-1}{\text{C}}_{r-1}$), and for 3 in 3 ways($b_1+b_2+b_3=1$). Hence, the answer is $3\times 10=30$.

The above assumes you are looking only for positive solutions. If $x,y,z$ are allowed to be negative, the solutions are either the ones found above (there are $30$ of these), or the ones found above with some two variables multiplied by $-1$ (there are $90$ of these). In total, we get $120$ solutions.

Hence option $C$ is correct