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Previous Year Question (PYQs)

The number of ordered pairs $(m,n)$, $m,n\in{1,2,\ldots,100}$ such that $7^m+7^n$ is divisible by $5$ is




Solution

$7\equiv2\pmod5$ So $7^k\equiv2^k\pmod5$ $2^k\pmod5$ cycles as $2,4,3,1$ (period $4$). $7^m+7^n\equiv0\pmod5$ when residues are complementary. Total valid ordered pairs $=2500$ Answer: $\boxed{2500}$

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