Question 62

What is the remainder when $$(127^{97} + 97^{97})$$ is divided by 32?

Solution
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$$(127^{97} + 97^{97})$$ is divided by 32 So,
$$\frac{(127^{97} + 97^{97})}{32}$$
=$$\frac{((128 - 1)^{97} + (96 + 1)^{97})}{32}$$
Remainder = -1 + 1 = 0

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