CAT 2019 Quant Question with Solution 65

Correct Answer: 13

Taking $2^{n d}$ equation

$5^{x-1}+3^{y+1}=9686$, the last digit of $5^{x-1}$ will always be 5 for all positive integral values of x

The power cycle of 3 is:

$3^{4 k+1} \equiv 3$

$3^{4 k+2} \equiv 9$

$3^{4 k+3} \equiv 7$

$3^{4 k} \equiv 1$

Clearly $3^{y+1}$ must be in the form of $3^{4 k}$ as the unit digit of R.H.S. =6

We have ${{3}^{4}}=81,\text{and}\ {{3}^{8}}=6561$

Also, $9686-81=9605$and $9686-6561=3125$

Observe that $3125=5^{5}$

Hence $5^{x-1}=5^{5}$

or $x=6$ and $3^{y+1}=3^{8} \Rightarrow y=7$

(x=6 and y=7 also satisfies the first equation)

Therefore, $x+y=6+7=13$