JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 22)
The number of elements in the set $\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\}$ equals __________.
Answer
169
Explanation
$$
x+2 y+3 z=42
$$
$$ x, y, z \geq 0 $$
as
$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\\\ z=5 & x+2 y=27 \Rightarrow 14 \text { cases }\end{array}$
$z=6 \quad x+2 y=24 \Rightarrow 13$ cases
$z=7 \quad x+2 y=21 \Rightarrow 11$ cases
$z=8 \quad x+2 y=18 \Rightarrow 10$ cases
$z=9 \quad x+2 y=15 \Rightarrow 8$ cases
$z=10 x+2 y=12 \Rightarrow 7$ cases
$z=11 x+2 y=9 \Rightarrow 5$ cases
$z=12 x+2 y=6 \Rightarrow 4$ cases
$z=13 x+2 y=3 \Rightarrow 2$ cases
$z=14 x+2 y=0 \Rightarrow 1$ case
Total = 169
$$ x, y, z \geq 0 $$
as
$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\\\ z=5 & x+2 y=27 \Rightarrow 14 \text { cases }\end{array}$
$z=6 \quad x+2 y=24 \Rightarrow 13$ cases
$z=7 \quad x+2 y=21 \Rightarrow 11$ cases
$z=8 \quad x+2 y=18 \Rightarrow 10$ cases
$z=9 \quad x+2 y=15 \Rightarrow 8$ cases
$z=10 x+2 y=12 \Rightarrow 7$ cases
$z=11 x+2 y=9 \Rightarrow 5$ cases
$z=12 x+2 y=6 \Rightarrow 4$ cases
$z=13 x+2 y=3 \Rightarrow 2$ cases
$z=14 x+2 y=0 \Rightarrow 1$ case
Total = 169
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