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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    @zepdrix hey man, do you have a second to help me on a problem?

  2. anonymous
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    @zepdrix Thanks again :)

  4. zepdrix
    • one year ago
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    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

  5. zepdrix
    • one year ago
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    yayaya that'll work nicely

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

  7. anonymous
    • one year ago
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    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) ??????????????????????????

  8. anonymous
    • one year ago
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    wait that's not right at all hold on.

  9. anonymous
    • one year ago
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    I feel like I need to factor out 25 and I'm not really quite sure how to do that.

  10. zepdrix
    • one year ago
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    AHHH sorry I have too many tabs open >.< I forgot about this one!

  11. zepdrix
    • one year ago
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    You rewrote 32 as (25+7)? Ok that seems good!

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

  13. zepdrix
    • one year ago
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    \[\large\rm 25\left(5^{2k}\right) \color{royalblue}{- (25+7)\left(2^{5k}\right)}\]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

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

  15. zepdrix
    • one year ago
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    On the last problem, we did something like this,|dw:1443323057344:dw|

  16. anonymous
    • one year ago
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    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.

  17. zepdrix
    • one year ago
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    ahh there we go! :) looks good!

  18. anonymous
    • one year ago
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    i don't know where i'd be without you zepdrix <3

  19. zepdrix
    • one year ago
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    yay team \c:/

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