Question

Total number of integral solutions of $|x|+|y|+|z|=15$ is $k$, then $\frac{k}{41}=$

Answer 1

Let $x,y,z$ be positive integers.
Then number of solutions = ${}^{14}{C}_2=91$
But $x,y,z$ can be negative too.
So, total number of solutions (none of $x,y,z$ is zero) is $91\times 2\times 2\times 2=728...\left(1\right)$

Let one of $x,y,z$ is equal to zero.
Then number of possible solutions $={}^{14}{C}_1=14$
But other two can be negative $(14\times 2\times 2)$ and one of three can be zero.
Hence, number of solutions $=14\times 2\times 2\times 3=168...\left(2\right)$

Let two of $x,y,z$ is zero, then number of solutions is $6...\left(3\right)$

Adding $\left(1\right),\left(2\right)$ and $\left(3\right)$
Total number of integral solutions, $k$ = 902
Hence, $\frac{k}{41}=\frac{902}{41}=22$