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Doubt in Modular Arithmetic
Today in 1557C - Moamen and XOR I wanted to compute $$${(2^{k-1})}^{n}$$$. But instead of writing binpow(binpow(2, k — 1), n), I simplified it to $$$2^{(k-1)*n}$$$ and wrote binpow(2, mul(k — 1, n)) and failed to pass pretest 2.
Why are not the two equivalent?
