Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

bbkzr31

Need help with a discrete math problem.

  • 5 months ago
  • 5 months ago

  • This Question is Closed
  1. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    • 5 months ago
    1 Attachment
  2. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    @myininaya

    • 5 months ago
  3. flyingace94
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 5 months ago
  4. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    so your saying 1 and 2 are the only positive integers which dived (n+1)!

    • 5 months ago
  5. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah, I understand the reasoning now. Thanks! :D

    • 5 months ago
  6. flyingace94
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 5 months ago
  7. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm, so would this be a proof by induction?

    • 5 months ago
  8. flyingace94
    Best Response
    You've already chosen the best response.
    Medals 1

    Induction should work.

    • 5 months ago
  9. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    right, just wondering if this is induction, because that was one of the hints for doing it.

    • 5 months ago
  10. flyingace94
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes, after two we can induce no other integers will work.

    • 5 months ago
  11. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    Just need you to check and see if this is the right way to work the problem.

    • 5 months ago
  12. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    I understand the thinking, I just want to ensure it is the right way to do it.

    • 5 months ago
  13. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 5 months ago
  14. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    If this helps, I have some notes on how I was suppose to start the problem.

    • 5 months ago
    1 Attachment
  15. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry, lost connection for a couple minutes.

    • 5 months ago
  16. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    So yes, proof by induction is how I was suppose to solve it

    • 5 months ago
  17. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    So by observing a pattern we do see see that each of those those quotients are between n and n-1

    • 5 months ago
  18. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    So that is how it appears in the pattern, but how would we actually prove it?

    • 5 months ago
  19. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    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...

    • 5 months ago
  20. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 5 months ago
  21. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    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)!)

    • 5 months ago
  22. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    So (k+2)!=(k+1)!*(k+2) And we know what (k+1)! is in between

    • 5 months ago
  23. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    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)?

    • 5 months ago
  24. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    still thinking...

    • 5 months ago
  25. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't know if this would help, something else I found on the problem. Like a start to a proof.

    • 5 months ago
    1 Attachment
  26. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    Keep losing connect, sorry -_-

    • 5 months ago
  27. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    Running to class now, I'll be back around 2.

    • 5 months ago
  28. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    \[k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)!\]

    • 5 months ago
  29. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 5 months ago
  30. bbkzr31
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • 5 months ago
  31. ganeshie8
    Best Response
    You've already chosen the best response.
    Medals 0

    \(k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)! \) \(=(k-1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)

    • 5 months ago
  32. ganeshie8
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 5 months ago
  33. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    -1(1!+...+k!)+2(1!+...+k!) = 1(1!+...+k!)

    • 5 months ago
  34. ganeshie8
    Best Response
    You've already chosen the best response.
    Medals 0

    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)...

    • 5 months ago
  35. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    I don't understant -1+2=1

    • 5 months ago
  36. ganeshie8
    Best Response
    You've already chosen the best response.
    Medals 0

    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 : \(=(k-1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)

    • 5 months ago
  37. ganeshie8
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 5 months ago
  38. myininaya
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah you are right

    • 5 months ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.