CAT 2019 Shift 2 Paper

Taking equation
$5^{\mathit{x}-1}+3^{\mathit{y}+1}=9686,$ the last digit of $5^{\mathit{x}-1}$ will always be 5 for all positive integral values of $\mathit{x}$
The power cycle of 3 is:
$3^{4\mathit{k}+1}\equiv 3$
$3^{4\mathit{k}+2}\equiv 9$
$3^{4\mathit{k}+3}\equiv 7$
$3^{4\mathit{k}}\equiv 1$
Clearly $3^{\mathit{y}+1}$ must be in the form of $3^{4\mathit{k}}$ as the unit digit of R.H.S.=6
We have $3^4=81,$ and $3^8=6561$
Also, $9686-81=9605$ and $9686-6561=3125$
Observe that $3125=5^5$
Hence $5^{\mathit{x}-1}=5^5$
or $\mathit{x}=6$ and $3^{\mathit{y}+1}=3^8\Rightarrow \mathit{y}=7$
$(\mathit{x}=6$ and $\mathit{y}=7$ also satisfies the first equation)
Therefore, $\mathit{x}+\mathit{y}=6+7=13$