The number of elements in the set\ $S=\{(x, y, z): x, y, z \in Z, x+2 y+3 z=42, x, y, z \geq 0\}$ e

Question

The number of elements in the set\ $S=\{(x, y, z): x, y, z \in Z, x+2 y+3 z=42, x, y, z \geq 0\}$ equals $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(169): $x+2 y+3 z=42$ At $z=0, x+2 y=42$\ $y$ can take all value from 0 to 21\ Hence 22 solutions.\ Similarly,\ at $z=1, x+2 y=39 \quad ightarrow \quad 20$ solutions\ at $z=2, x+2 y=36 \quad ightarrow \quad 19$ solutions\ at $z=3, x+2 y=33 ightarrow 17$ solutions\ at $z=4, x+2 y=30 ightarrow 16$ solutions\ at $z=5, x+2 y=27 \quad ightarrow \quad 14$ solutions\ at $z=6, x+2 y=24 \quad ightarrow \quad 13$ solutions\ at $z=7, x+2 y=21 \quad ightarrow \quad 11$ solutions\ at $z=8, x+2 y=18 \quad ightarrow \quad 10$ solutions\ at $z=9, x+2 y=15 \quad ightarrow \quad 8$ solutions\ at $z=10, x+2 y=12 \quad ightarrow \quad 7$ solutions\ at $z=11, x+2 y=9 \quad ightarrow \quad 5$ solutions\ at $z=12, x+2 y=6 \quad ightarrow \quad 4$ solutions\ at $z=13, x+2 y=3 \quad ightarrow \quad 2$ solutions\ at $z=14, x+2 y=0 \quad ightarrow \quad 1$ solutions\ Total cases $=169$ solutions.\