Prove that the last two digits are 28

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 $28$ .

#NumberTheory #Finn #Casework #CanYouProveIt?

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Hint:Induction

Bogdan Simeonov - 7 years, 1 month ago

Okay... Can you explain?

Finn Hulse - 7 years, 1 month ago

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 - 7 years, 1 month ago

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

Finn Hulse - 7 years, 1 month ago

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

Bogdan Simeonov - 7 years, 1 month ago

@Bogdan Simeonov – Probably. How? :D

Finn Hulse - 7 years, 1 month ago

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

Bogdan Simeonov - 7 years, 1 month ago

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

Finn Hulse - 7 years, 1 month ago

@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 - 7 years, 1 month ago

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

Finn Hulse - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$