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Prove that if \(n\) is an odd positive integer, then the last two digits of \(2^{2n}(2^{2n+1}−1)\) in base \(10\) are \(28\).

Note by Finn Hulse 3 years, 5 months ago

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Hint:Induction

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Okay... Can you explain?

When n=1, the result is 28.Suppose for some n the condition is true.

Multiply by 16^2 and add \(15.2^{2n+4}\) and you will get the next such number.So the new number will be congruent to

\(256.28+15.2^{2n+4}\equiv 68+15.2^{2n+4}(mod100)\)

But \(15.2^{2n}\) for odd numbers n will be congruent to 60 modulo 100 (that's not hard to see), so we have proven in inductively.

@Bogdan Simeonov – Oh, I guess so! Great job Bogdan. :D

@Finn Hulse – You probably have no idea how to pronounce my name :D

@Bogdan Simeonov – Probably. How? :D

@Finn Hulse – Try using the speech feature on it.It sounds stupid :D

@Bogdan Simeonov – HAHA yeah. But how is your name actually pronounced?

@Finn Hulse – I believe it should be pronounced like this: pronounce log, but with a B instead of an l, then say the name Dan :D

@Bogdan Simeonov – Oh, that's what I thought. :D

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TopNewestHint:Induction

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Okay... Can you explain?

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When n=1, the result is 28.Suppose for some n the condition is true.

Multiply by 16^2 and add \(15.2^{2n+4}\) and you will get the next such number.So the new number will be congruent to

\(256.28+15.2^{2n+4}\equiv 68+15.2^{2n+4}(mod100)\)

But \(15.2^{2n}\) for odd numbers n will be congruent to 60 modulo 100 (that's not hard to see), so we have proven in inductively.

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