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- anonymous

a

- schrodinger

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- anonymous

@zepdrix hey man, do you have a second to help me on a problem?

- anonymous

Prove that for all integers n > 0, 5^2n – 2^5n is divisible by 7.
Base Case: n=1
5^(2(1))-2^(5(1)) = 5^2-2^5 = 25-32 = -7
-7|7
Induction Hypothesis: n=k
5^(2(k+1) – 2^(5(k+1)
5^(2k+2) - 2^(5k+5)
5^(2k) * 5^2 - 2^5k * 2^5
25(5^(2k)) - 32(2^(5k))
So close. How do I get 5^2n – 2^5n ... to solve the equation?

- anonymous

@zepdrix
Thanks again :)

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## More answers

- zepdrix

Hmm I think it's going to be the same sneaky trick that we applied last time, using the base case.
Sec, I'm working it on paper to make sure :D

- zepdrix

yayaya that'll work nicely

- zepdrix

\[\large\rm 25\left(5^{2k}\right) - \color{orangered}{32}\left(2^{5k}\right)\]Let's rewrite this 32 in another fancy way.
Look back at your base case to get some ideas :d

- anonymous

I'm going to end up with:
25(5^(2k)) - (25+7)(2^(5k))
25(5^(2k)) - (7)(2^(5k))
25(5^(2k)) - (2^(5k))(7)
??????????????????????????

- anonymous

wait that's not right at all hold on.

- anonymous

I feel like I need to factor out 25 and I'm not really quite sure how to do that.

- zepdrix

AHHH sorry I have too many tabs open >.< I forgot about this one!

- zepdrix

You rewrote 32 as (25+7)?
Ok that seems good!

- zepdrix

\[\large\rm 25\left(5^{2k}\right) - \color{orangered}{32}\left(2^{5k}\right)\]\[\large\rm 25\left(5^{2k}\right) - \color{orangered}{(25+7)}\left(2^{5k}\right)\]Are you having trouble with this next step again?
Remember distributing? :)

- zepdrix

\[\large\rm 25\left(5^{2k}\right) \color{royalblue}{- (25+7)\left(2^{5k}\right)}\]Distributing,\[\large\rm 25\left(5^{2k}\right)\color{royalblue}{ - 25\left(2^{5k}\right)-7\left(2^{5k}\right)}\]That, ya?
Gotta remember to distribute the negative sign as well :o

- anonymous

Yeah, I've got that written down on my paper. I'm just trying to compare it to the last problem we did.

- zepdrix

On the last problem, we did something like this,|dw:1443323057344:dw|

- anonymous

yeah, that's why i just did on my paper. honest, i got it before you posted that. we factored out the 8. i was a little hesitant here. this isn't, like, typical factoring set up.
the way i think about it is the number i factor out basically conjoins the terms. seems like that sentiment is confirmed, here. i wrote it 25(5^k2-2^5k)-(2^5k)7 and that should complete the proof. we've got 25x the assumption and the right is multiplied by 7... which implies it can be divided by 7.

- zepdrix

ahh there we go! :) looks good!

- anonymous

i don't know where i'd be without you zepdrix <3

- zepdrix

yay team \c:/

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