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anonymous
 one year ago
a
anonymous
 one year ago
a

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix hey man, do you have a second to help me on a problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Prove that for all integers n > 0, 5^2n – 2^5n is divisible by 7. Base Case: n=1 5^(2(1))2^(5(1)) = 5^22^5 = 2532 = 7 77 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
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix Thanks again :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm 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
 one year ago
Best ResponseYou've already chosen the best response.1yayaya that'll work nicely

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm 25\left(5^{2k}\right)  \color{orangered}{32}\left(2^{5k}\right)\]Let's rewrite this 32 in another fancy way. Look back at your base case to get some ideas :d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0wait that's not right at all hold on.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I feel like I need to factor out 25 and I'm not really quite sure how to do that.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1AHHH sorry I have too many tabs open >.< I forgot about this one!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You rewrote 32 as (25+7)? Ok that seems good!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm 25\left(5^{2k}\right)  \color{orangered}{32}\left(2^{5k}\right)\]\[\large\rm 25\left(5^{2k}\right)  \color{orangered}{(25+7)}\left(2^{5k}\right)\]Are you having trouble with this next step again? Remember distributing? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm 25\left(5^{2k}\right) \color{royalblue}{ (25+7)\left(2^{5k}\right)}\]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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I've got that written down on my paper. I'm just trying to compare it to the last problem we did.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1On the last problem, we did something like this,dw:1443323057344:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, 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^k22^5k)(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
 one year ago
Best ResponseYou've already chosen the best response.1ahh there we go! :) looks good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't know where i'd be without you zepdrix <3
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