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anonymous

  • one year ago

Prove that for all integers n>0, 5^(2n)-2^(5n) is divisible by 7.

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  1. amistre64
    • one year ago
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    sounds like some sort of induction process to me

  2. anonymous
    • one year ago
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    It is an induction problem. I am typing out the work I have down, now.

  3. anonymous
    • one year ago
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    Base Case: 5^((2)(1))-2^((5)(1)) is divisible by 7. 5^(2)*5-2^(5)*2=61 61|7 Inductive Hypothesis: n=k 5^((2)(k+1))-2^((5)(k+1)) 5^((2)(k)*5-2^((5)(k))*2

  4. amistre64
    • one year ago
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    i think your induction is off

  5. anonymous
    • one year ago
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    I 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.

  6. amistre64
    • one year ago
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    5^(2(k+1)) - 2^(5(k+1)) 5^(2k +2) - 2^(5k +5) 5^(2k) 5^2 - 2^(5k) 2^5

  7. anonymous
    • one year ago
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    Okay, let me try to work that out. I'll post what I have in couple moments.

  8. amistre64
    • one year ago
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    5^(2n) - 2^(5n) [5^2]^n - [2^5]^n this might be a better format to play with

  9. anonymous
    • one year ago
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    @AlexADB definitely got the right idea

  10. amistre64
    • one year ago
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    5^(2(k+1)) - 2^(5(k+1)) [5^2]^(k+1) -[2^5]^(k+1)

  11. amistre64
    • one year ago
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    but yeah, ive got no clear direction for this in my head yet

  12. anonymous
    • one year ago
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    well, that was annoying. i had a problem posting for a second. okay, still trying to figure this out.

  13. anonymous
    • one year ago
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    take out the 25 might work

  14. anonymous
    • one year ago
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    \[25\times 5^{2k}-32\times 2^{5k}\]

  15. anonymous
    • one year ago
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    \[25(5^{2k}- 2^{5k})-7\times 2^{5k}\]

  16. anonymous
    • one year ago
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    seen it before is how i know

  17. amistre64
    • one year ago
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    i was trying to divide it by 5^k - 2^k to see where it led :)

  18. anonymous
    • one year ago
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    i 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

  19. amistre64
    • one year ago
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    32 2^(5k) = 25 2^(5k) +7 2^(5k)

  20. amistre64
    • one year ago
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    25 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

  21. amistre64
    • one year ago
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    both sides are factorable by 7, and therefore ...

  22. amistre64
    • one year ago
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    both *terms* are factorable by 7 that is

  23. amistre64
    • one year ago
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    does it make sense yet?

  24. anonymous
    • one year ago
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    im copying that onto paper. i need another 2 minutes to look at it.

  25. amistre64
    • one year ago
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    25x + 32y 32y = 25y + 7y 25x + 25y + 7y 25(x + y) + 7y but it was assumed that 7|x+y 25*7s + 7y 7(25s + y)

  26. amistre64
    • one year ago
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    not sure if thats the proper method, but its proof enough to me

  27. anonymous
    • one year ago
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    Base Case: 5^((2)(1))-2^((5)(1)) is divisible by 7. 5^(2)*5-2^(5)*2=61 61|7 Inductive Hypothesis: n=k 5^((2)(k+1))-2^((5)(k+1)) 5^((2)(k)*5-2^((5)(k))*2 Okay, I kind of need to see this from the beginning. First, is my base case and inductive hypothesis right?

  28. amistre64
    • one year ago
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    your induction is off again ...

  29. amistre64
    • one year ago
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    might help to get rid of the 'bases' and use the integer representations

  30. anonymous
    • one year ago
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    Which 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?'

  31. amistre64
    • one year ago
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    for example ... 5^((2)(k+1)) is not 5^(2k) * 5

  32. amistre64
    • one year ago
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    5^(2k) is a base of 5, to a power if 2k ... 5^(2k) = 25^k

  33. anonymous
    • one year ago
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    i 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?

  34. amistre64
    • one year ago
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    5^(2(k+1)) = 5^(2k+2) = 5^(2k) * 5^2 you are not distributing correctly and it is throwing off your setup

  35. mathmate
    • one year ago
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    Alternatively, as you did it: \(S(n)=5^{2n}-2^{5n}\) => Base case: n=1 \( S(1)=5^{2\times 1}-2^{5\times 1}=25-32=-7\) => 7|S(1)

  36. anonymous
    • one year ago
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    @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)*5-2^(5k)*2 After this, I am not quite sure what to do. I want 5^2k-2^5k because that assumption is being treated as true. But, what should be done with the 5 and 2 that are in the way?

  37. mathmate
    • one year ago
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    A lot of your misery is in mis-distribution. 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 7|S(n) For the Induction step, You have to show that 7|S(n+1), \(given\) that 7|S(n). Re-read @Amistre64's last few posts for hints on this step.

  38. anonymous
    • one year ago
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    @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.

  39. amistre64
    • one year ago
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    Lets 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 (x-y) we can just as easily add zero to this setup,say (7x-7x) 25x + (0) - 32y 25x + (7x-7x) - 32y 25x + 7x -7x - 32y (25x + 7x) -7x - 32y 32x -7x - 32y 32x - 32y - 7x (32x - 32y) - 7x 32(x - y) - 7x

  40. amistre64
    • one year ago
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    it is assumed, or given, that (x-y) is divisible by 7 ... so it is some value 7s.

  41. zepdrix
    • one year ago
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    waow that's some pretty LaTeX :) lol underbrace? Hmm never seen that before :3

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