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anonymous

  • one year ago

Prove that for all integers n > 0, 8n – 3n is divisible by 5. check my answer?

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  1. anonymous
    • one year ago
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    Base Case: n=1 8^1-3^1=5 5/5=1 Induction Hypothesis: 8^k-3^k is divisible by 3 8^(k+1)-3^(k+1) 8^(2)-3^(2) 64-9=53 53/3

  2. jim_thompson5910
    • one year ago
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    I think you meant to say `Induction Hypothesis: 8^k-3^k is divisible by 5`

  3. jim_thompson5910
    • one year ago
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    if we assume that claim is true, then we can say 8^k-3^k = 5q where q is some integer. Agreed?

  4. anonymous
    • one year ago
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    what is the significance of 5q??

  5. jim_thompson5910
    • one year ago
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    it just means "5 times some integer"

  6. jim_thompson5910
    • one year ago
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    if I said "100 is divisible by 5" I can state that as 100 = 5q In this case, q = 20

  7. anonymous
    • one year ago
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    ok, yeah. so that just means divisible by 5

  8. jim_thompson5910
    • one year ago
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    yeah

  9. jim_thompson5910
    • one year ago
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    the goal is to prove that if 8^k-3^k = 5q is assumed to be true, then 8^(k+1)-3^(k+1) = 5s where q and s are integers

  10. jim_thompson5910
    • one year ago
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    here are 2 hints: 1) isolate the `8^k` in `8^k-3^k = 5q` to get `8^k = 5q + 3^k` 2) break down `8^(k+1)-3^(k+1) ` into `8*8^k-3*3^k`

  11. anonymous
    • one year ago
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    i do not understand.8^(k+1) is just 64, no? not 64^K?

  12. jim_thompson5910
    • one year ago
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    how are you getting 64? you can't replace k with 1

  13. anonymous
    • one year ago
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    i didn't know i couldn't do that.

  14. jim_thompson5910
    • one year ago
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    you can replace k with 1 if you wanted to prove that specific piece. But we're trying to prove the whole thing in general (for any number k)

  15. jim_thompson5910
    • one year ago
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    do you see how I turned `8^(k+1)-3^(k+1)` into `8*8^k-3*3^k` ?

  16. anonymous
    • one year ago
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    no, that was my next question. i thought k+1 would turn into ^2

  17. jim_thompson5910
    • one year ago
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    recall that \[\LARGE x^{y+z} = x^y * x^z\]

  18. jim_thompson5910
    • one year ago
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    so \[\Large 8^{k+1} = 8^k*8^1 = 8^k*8 = 8*8^k\] make sense?

  19. anonymous
    • one year ago
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    yes. thank you. and now i have to do the same thing for the other side, right?

  20. jim_thompson5910
    • one year ago
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    for the 3^(k+1) part you mean? yes

  21. anonymous
    • one year ago
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    ok, so i just get 3^k*3+5q=6k+5q

  22. jim_thompson5910
    • one year ago
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    we have 8*8^k-3*3^k now replace the 8^k with 5q + 3^k 8*8^k-3*3^k 8*(5q + 3^k) - 3*3^k 8*5q + 8*3^k - 3*3^k do you see what's next?

  23. anonymous
    • one year ago
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    multiply

  24. anonymous
    • one year ago
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    40q + 24^k - 9^k 40q + 15^k

  25. jim_thompson5910
    • one year ago
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    8*3^k isn't equal to 24^k

  26. jim_thompson5910
    • one year ago
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    8*5q + 8*3^k - 3*3^k 5*8q + 8*3^k - 3*3^k 5*8q + (8*3^k - 3*3^k) 5*8q + (8 - 3)*3^k 5*8q + 5*3^k 5*(8q + 3^k) 5*s where s = 8q + 3^k and s is an integer. So this all proves that if 8^k - 3^k = 5q, then 8^(k+1)-3^(k+1) = 5s q and s are integers

  27. anonymous
    • one year ago
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    im going to have to study that, tomorrow. can you look at a different kind of problem with me?

  28. jim_thompson5910
    • one year ago
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    sure

  29. anonymous
    • one year ago
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    1^3 + 2^3 + 3^3 + … + n^3 = n^2(n + 1)^2/4 for all positive integers n

  30. jim_thompson5910
    • one year ago
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    I'm sure you have the base case already done?

  31. anonymous
    • one year ago
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    no. i just started looking at it. i have not tried solving a sequence kind of problem yet.

  32. anonymous
    • one year ago
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    n^2(n^2+1)/4

  33. jim_thompson5910
    • one year ago
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    you'll use induction here

  34. anonymous
    • one year ago
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    i have less idea about this problem then i did the last :|

  35. jim_thompson5910
    • one year ago
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    the base case is n = 1

  36. jim_thompson5910
    • one year ago
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    prove that is true

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