## anonymous one year ago a

1. anonymous

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

2. 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?

3. anonymous

@zepdrix Thanks again :)

4. 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

5. zepdrix

yayaya that'll work nicely

6. 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

7. 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) ??????????????????????????

8. anonymous

wait that's not right at all hold on.

9. anonymous

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

10. zepdrix

11. zepdrix

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

12. 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? :)

13. 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

14. anonymous

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

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

16. 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.

17. zepdrix

ahh there we go! :) looks good!

18. anonymous

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

19. zepdrix

yay team \c:/