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2011/12 AHSMC Part 1 Problem 1

Problem 1 of 16

If 22012+41006=2n2^{2012} + 4^{1006} = 2^n, then nn is

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Solution

We have 22012+41006=22012+22012=220132^{2012} + 4^{1006} = 2^{2012} + 2^{2012} = 2^{2013}. The answer is (A).

University of Alberta