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Need help with a discrete math problem.

Mathematics
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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.

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Other answers:

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

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