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bbkzr31

  • one year ago

Need help with a discrete math problem.

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  1. bbkzr31
    • one year ago
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  2. bbkzr31
    • one year ago
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    @myininaya

  3. flyingace94
    • one year ago
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    1,2. the following might not be right but it helped me get my answer: 2I6 or 2!I3!, 6I24 or 3!I4!, 24I120 or 4!I5!, I see that n!I(n+1)! Is there a formula that relates the sum of factorials? speaking intuitively I see that 1!+2!+...+n! is always odd as every factorial above 2! is a multiple of two and an even number plus 1 is always odd. Using this I can see that n cannot be greater than 2. so n=1,2.

  4. bbkzr31
    • one year ago
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    so your saying 1 and 2 are the only positive integers which dived (n+1)!

  5. bbkzr31
    • one year ago
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    Ah, I understand the reasoning now. Thanks! :D

  6. flyingace94
    • one year ago
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    I just guessed and check on 1 and 2 but no other integers seem to work, i check 3-7 and found that they didn't work. Sorry, I've never take discrete math.

  7. bbkzr31
    • one year ago
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    Hmm, so would this be a proof by induction?

  8. flyingace94
    • one year ago
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    Induction should work.

  9. bbkzr31
    • one year ago
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    right, just wondering if this is induction, because that was one of the hints for doing it.

  10. flyingace94
    • one year ago
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    Yes, after two we can induce no other integers will work.

  11. bbkzr31
    • one year ago
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    Just need you to check and see if this is the right way to work the problem.

  12. bbkzr31
    • one year ago
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    I understand the thinking, I just want to ensure it is the right way to do it.

  13. myininaya
    • one year ago
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    n=3 is the first case we notice both sides aren't equal... We could attempt by induction to prove that (1!+2!+...+n!) does not divide (n+1)! for n>=3.

  14. bbkzr31
    • one year ago
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    If this helps, I have some notes on how I was suppose to start the problem.

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  15. bbkzr31
    • one year ago
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    sorry, lost connection for a couple minutes.

  16. bbkzr31
    • one year ago
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    So yes, proof by induction is how I was suppose to solve it

  17. myininaya
    • one year ago
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    So by observing a pattern we do see see that each of those those quotients are between n and n-1

  18. bbkzr31
    • one year ago
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    So that is how it appears in the pattern, but how would we actually prove it?

  19. myininaya
    • one year ago
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    So we have n=3 is like totally true based on your notebook paper. That step is already done. 2<24/9<3 is a totally true statement we need to assume that for some integer k>=3 we have \[k-1<\frac{(k+1)!}{1!+2!+...+k!}<k\] We to show the following expression is true: \[(k+1)-1<\frac{((k+1)+1)!}{1!+2!+...+k!+(k+1)!}<k+1\] and hmmm.... thinking...

  20. myininaya
    • one year ago
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    We know that (k+2)!=(k+1)! * (k+2) But I don't know what I'm gonna do what that bottom yet...

  21. myininaya
    • one year ago
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    I wonder if it would be easier to look at: (k-1)(1!+2!+...+k!)<(k+1)!<(k)(1!+2!+...+k!) and showing the following: k(1!+2!+...+(k+1)!)<(k+2)!<(k+1)(1!+2!+...+(k+1)!)

  22. myininaya
    • one year ago
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    So (k+2)!=(k+1)!*(k+2) And we know what (k+1)! is in between

  23. myininaya
    • one year ago
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    So if we look at our expression that we assumed is true what do we get when we multiply all sides of the inequality by (k+2)?

  24. myininaya
    • one year ago
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    still thinking...

  25. bbkzr31
    • one year ago
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    I don't know if this would help, something else I found on the problem. Like a start to a proof.

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  26. bbkzr31
    • one year ago
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    Keep losing connect, sorry -_-

  27. bbkzr31
    • one year ago
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    Running to class now, I'll be back around 2.

  28. myininaya
    • one year ago
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    \[k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)!\]

  29. myininaya
    • one year ago
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    \[=(k-1)(1!+...+k!)+k(k+1)! +2(1!+...+k!) \] \[\le (k+1)!+k(k+1)!+2(1!+...+k!) \] \[=(k+1)!(k+1)+2(1!+...+k!)\] \[\le (k+1)!(k+1)+(k+1)!=(k+1)!(k+1+1)=(k+2)!\] This should help... Now maybe you can prove the other side.

  30. bbkzr31
    • one year ago
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    Awesome, thank you so much for the help! :) Also, I asked my teacher about the recursive problem you helped me with were we were thinking the answer was 0. He told me I could just leave the answer as A(1,A(2,2^16-1)) which is exactly what you had gotten.

  31. ganeshie8
    • one year ago
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    \(k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)! \) \(=(k-1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)

  32. ganeshie8
    • one year ago
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    im trying to understand the proof, @myininaya im thinking that 2 should not be there ? i can see it wont change the proof... as 1, 2 are both < k, when k>3. just im wondering if that 2 is a typo or im missing something

  33. myininaya
    • one year ago
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    -1(1!+...+k!)+2(1!+...+k!) = 1(1!+...+k!)

  34. ganeshie8
    • one year ago
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    sorry i get that before. it doesnt clear my doubt. my concern is, 1(1!+...+k!) is coming as extra in second line above (my reply)...

  35. myininaya
    • one year ago
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    I don't understant -1+2=1

  36. ganeshie8
    • one year ago
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    sorry to persist lol.. im really not getting it... i just want to knw how to get to step2 from step 1 step 1 : \(k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)!\) step 2 : \(=(k-1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)

  37. ganeshie8
    • one year ago
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    i thought step2 should be :- step2 : \(=(k-1)(1!+...+k!)+k(k+1)! +\color{red}{1}(1!+...+k!)\) that 2 in ur step2 is throwing me off

  38. myininaya
    • one year ago
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    yeah you are right

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