SSC CGL 17 Aug 2021 Shift 1 Paper

Given, Number $94\mathit{x}29$ y 6 is divisible by 72 then it is also divisible by 8 and 9 (factors of 72 ) divisibility of 8 , last 3 digits of the number should be divisibile by 8
$\therefore$ Last 3 -digits of number $=9\mathit{y}6$ for $\mathit{y}=3,7,9\mathit{y}6$ is divisible by 8
$(\because 936÷8=117$ and $976÷8=122)$ divisibility of 9
Sum of digits of the number should be divisible by 9 .
$\therefore$ Sum of digits
$=9+4+\mathit{x}+2+9+\mathit{y}+6$
$=30+\mathit{x}+\mathit{y}$
for $\mathit{y}=3$, then $\mathit{x}=3$.
$=30+3+3=36$
(which is divisible by 9 )
for $\mathit{y}=7$, then $\mathit{x}=8$
$=30+8+7=45$
(which is divisible by 9 )
According to the question,
$\mathit{x}\ne \mathit{y}$

$\therefore 2\mathit{x}+3\mathit{y}=2\times 8+3\times 7$
$=16+21=37$