Question

a

Answer 1

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

Answer 2

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?

Answer 3

@zepdrix

Thanks again :)

Answer 4

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

Answer 5

yayaya that'll work nicely

Answer 6

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

Answer 7

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)
??????????????????????????

Answer 8

wait that's not right at all hold on.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

\[\large\rm 25\left(5^{2k}\right) \color{royalblue}{- (25+7)\left(2^{5k}\right)}\]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

Answer 14

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

Answer 15

On the last problem, we did something like this,