Question:
If $5^{x}-3^{y}=13438$ and $5^{x-1}+3^{y+1}=9686,$ then x+y equals
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, $x+y=6+7=13$
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