anonymous
  • anonymous
Need help with a discrete math problem.
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so your saying 1 and 2 are the only positive integers which dived (n+1)!
anonymous
  • anonymous
Ah, I understand the reasoning now. Thanks! :D
anonymous
  • anonymous
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.
anonymous
  • anonymous
Hmm, so would this be a proof by induction?
anonymous
  • anonymous
Induction should work.
anonymous
  • anonymous
right, just wondering if this is induction, because that was one of the hints for doing it.
anonymous
  • anonymous
Yes, after two we can induce no other integers will work.
anonymous
  • anonymous
Just need you to check and see if this is the right way to work the problem.
anonymous
  • anonymous
I understand the thinking, I just want to ensure it is the right way to do it.
myininaya
  • myininaya
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.
anonymous
  • anonymous
If this helps, I have some notes on how I was suppose to start the problem.
1 Attachment
anonymous
  • anonymous
sorry, lost connection for a couple minutes.
anonymous
  • anonymous
So yes, proof by induction is how I was suppose to solve it
myininaya
  • myininaya
So by observing a pattern we do see see that each of those those quotients are between n and n-1
anonymous
  • anonymous
So that is how it appears in the pattern, but how would we actually prove it?
myininaya
  • myininaya
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!}
myininaya
  • myininaya
We know that (k+2)!=(k+1)! * (k+2) But I don't know what I'm gonna do what that bottom yet...
myininaya
  • myininaya
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)!)
myininaya
  • myininaya
So (k+2)!=(k+1)!*(k+2) And we know what (k+1)! is in between
myininaya
  • myininaya
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)?
myininaya
  • myininaya
still thinking...
anonymous
  • anonymous
I don't know if this would help, something else I found on the problem. Like a start to a proof.
1 Attachment
anonymous
  • anonymous
Keep losing connect, sorry -_-
anonymous
  • anonymous
Running to class now, I'll be back around 2.
myininaya
  • myininaya
\[k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)!\]
myininaya
  • myininaya
\[=(k-1)(1!+...+k!)+k(k+1)! +2(1!+...+k!) \] \[\le (k+1)!+k(k+1)!+2(1!+...+k!) \] \[=(k+1)!(k+1)+2(1!+...+k!)\] \[\le (k+1)!(k+1)+(k+1)!=(k+1)!(k+1+1)=(k+2)!\] This should help... Now maybe you can prove the other side.
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
\(k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)! \) \(=(k-1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)
ganeshie8
  • ganeshie8
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
myininaya
  • myininaya
-1(1!+...+k!)+2(1!+...+k!) = 1(1!+...+k!)
ganeshie8
  • ganeshie8
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)...
myininaya
  • myininaya
I don't understant -1+2=1
ganeshie8
  • ganeshie8
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 : \(=(k-1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)
ganeshie8
  • ganeshie8
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
myininaya
  • myininaya
yeah you are right

Looking for something else?

Not the answer you are looking for? Search for more explanations.