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anonymous
 one year ago
Prove that for all integers n>0, 5^(2n)2^(5n) is divisible by 7.
anonymous
 one year ago
Prove that for all integers n>0, 5^(2n)2^(5n) is divisible by 7.

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.3sounds like some sort of induction process to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is an induction problem. I am typing out the work I have down, now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Base Case: 5^((2)(1))2^((5)(1)) is divisible by 7. 5^(2)*52^(5)*2=61 617 Inductive Hypothesis: n=k 5^((2)(k+1))2^((5)(k+1)) 5^((2)(k)*52^((5)(k))*2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3i think your induction is off

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am not sure how to proceed from here. I know that I need to get 5^((2)(k))2^((5)(k) because we have assumed that is divisible. Then the rest needs to be proved  by 7.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.35^(2(k+1))  2^(5(k+1)) 5^(2k +2)  2^(5k +5) 5^(2k) 5^2  2^(5k) 2^5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, let me try to work that out. I'll post what I have in couple moments.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.35^(2n)  2^(5n) [5^2]^n  [2^5]^n this might be a better format to play with

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@AlexADB definitely got the right idea

amistre64
 one year ago
Best ResponseYou've already chosen the best response.35^(2(k+1))  2^(5(k+1)) [5^2]^(k+1) [2^5]^(k+1)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3but yeah, ive got no clear direction for this in my head yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, that was annoying. i had a problem posting for a second. okay, still trying to figure this out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0take out the 25 might work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[25\times 5^{2k}32\times 2^{5k}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[25(5^{2k} 2^{5k})7\times 2^{5k}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0seen it before is how i know

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3i was trying to divide it by 5^k  2^k to see where it led :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am trying to piece together the steps you guys are throwing around. i'm not good at math so i'm having an issue keeping up :X

amistre64
 one year ago
Best ResponseYou've already chosen the best response.332 2^(5k) = 25 2^(5k) +7 2^(5k)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.325 5^(2k)  (32 2^(5k)) 25 5^(2k)  (25 2^(5k) +7 2^(5k)) 25 5^(2k)  25 2^(5k) 7 2^(5k) [25 5^(2k)  25 2^(5k)] 7 2^(5k) etc ... it was not obvious to me at first either

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3both sides are factorable by 7, and therefore ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3both *terms* are factorable by 7 that is

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3does it make sense yet?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im copying that onto paper. i need another 2 minutes to look at it.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.325x + 32y 32y = 25y + 7y 25x + 25y + 7y 25(x + y) + 7y but it was assumed that 7x+y 25*7s + 7y 7(25s + y)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3not sure if thats the proper method, but its proof enough to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Base Case: 5^((2)(1))2^((5)(1)) is divisible by 7. 5^(2)*52^(5)*2=61 617 Inductive Hypothesis: n=k 5^((2)(k+1))2^((5)(k+1)) 5^((2)(k)*52^((5)(k))*2 Okay, I kind of need to see this from the beginning. First, is my base case and inductive hypothesis right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3your induction is off again ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3might help to get rid of the 'bases' and use the integer representations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which part of my induction is wrong? Is it the base case? Is it the Hypothesis? What do you mean by getting rid of the 'bases?'

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3for example ... 5^((2)(k+1)) is not 5^(2k) * 5

amistre64
 one year ago
Best ResponseYou've already chosen the best response.35^(2k) is a base of 5, to a power if 2k ... 5^(2k) = 25^k

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i didn't know i couldn't do that. 5^2 is the same as 5x5. I just reach up and grab a 5 from the exponent when I need more stuff to make the equation work, no?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.35^(2(k+1)) = 5^(2k+2) = 5^(2k) * 5^2 you are not distributing correctly and it is throwing off your setup

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Alternatively, as you did it: \(S(n)=5^{2n}2^{5n}\) => Base case: n=1 \( S(1)=5^{2\times 1}2^{5\times 1}=2532=7\) => 7S(1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate Thank you. I like your notation mathmate. Next, my Inductive Hypothesis would be: 5^(2(k+1))2^(5(k+1)) I don't think there is any trouble there. Inductive Step: 5^(2k)*52^(5k)*2 After this, I am not quite sure what to do. I want 5^2k2^5k because that assumption is being treated as true. But, what should be done with the 5 and 2 that are in the way?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0A lot of your misery is in misdistribution. Once that's fixed, you can go along with what @Amistre64 suggested. For the Inductive Hypothesis, it's just for the general case "n", or \( 7~~(5^{2n}2^{5n})\), or simply 7S(n) For the Induction step, You have to show that 7S(n+1), \(given\) that 7S(n). Reread @Amistre64's last few posts for hints on this step.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 hey man, I still don't understand what is going on in this problem. I understand 5^2k * 5^2  2^5k * 2^5; so, doing 25 and 32 i understand. but, getting here 25(5^2k−2^5k)−7×2^5k i don't see it. If you have time that would be awesome.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3Lets just clean up the expression \[\Huge 25(\underbrace{5^{2k}}_{\color{red}{=x}})32(\underbrace{2^{5k}}_{\color{green}{=y}})\] Now, we have 25x  32y ... and we want to factor out (xy) we can just as easily add zero to this setup,say (7x7x) 25x + (0)  32y 25x + (7x7x)  32y 25x + 7x 7x  32y (25x + 7x) 7x  32y 32x 7x  32y 32x  32y  7x (32x  32y)  7x 32(x  y)  7x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3it is assumed, or given, that (xy) is divisible by 7 ... so it is some value 7s.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0waow that's some pretty LaTeX :) lol underbrace? Hmm never seen that before :3
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