The number of elements in the set S=(x,y,z):x,y,z Z,x+2y+3z=42,x,y,z 0 equals __________.

Question

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

Answer: 169

Accepted Answer

$x+2y+3z=42$

$x,y,z\ge 0$

as

$\begin{gathered} z=0 x+2y=42\Rightarrow 22 \ \ z=1 x+2y=39\Rightarrow 20 \ \ z=2 x+2y=36\Rightarrow 19 \ \ z=3 x+2y=33\Rightarrow 17 \ \ z=4 x+2y=30\Rightarrow 16 \ \ z=5 x+2y=27\Rightarrow 14 \end{gathered}$

$z=6\, x+2y=24\Rightarrow 13$ cases

$z=7\, x+2y=21\Rightarrow 11$ cases

$z=8\, x+2y=18\Rightarrow 10$ cases

$z=9\, x+2y=15\Rightarrow 8$ cases

$z=10x+2y=12\Rightarrow 7$ cases

$z=11x+2y=9\Rightarrow 5$ cases

$z=12x+2y=6\Rightarrow 4$ cases

$z=13x+2y=3\Rightarrow 2$ cases

$z=14x+2y=0\Rightarrow 1$ case

Total = 169

Solutions · 1