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flyingace94
 one year ago
Best ResponseYou've already chosen the best response.11,2. 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.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0so your saying 1 and 2 are the only positive integers which dived (n+1)!

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0Ah, I understand the reasoning now. Thanks! :D

flyingace94
 one year ago
Best ResponseYou've already chosen the best response.1I just guessed and check on 1 and 2 but no other integers seem to work, i check 37 and found that they didn't work. Sorry, I've never take discrete math.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, so would this be a proof by induction?

flyingace94
 one year ago
Best ResponseYou've already chosen the best response.1Induction should work.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0right, just wondering if this is induction, because that was one of the hints for doing it.

flyingace94
 one year ago
Best ResponseYou've already chosen the best response.1Yes, after two we can induce no other integers will work.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0Just need you to check and see if this is the right way to work the problem.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0I understand the thinking, I just want to ensure it is the right way to do it.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1n=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.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0If this helps, I have some notes on how I was suppose to start the problem.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0sorry, lost connection for a couple minutes.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0So yes, proof by induction is how I was suppose to solve it

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1So by observing a pattern we do see see that each of those those quotients are between n and n1

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0So that is how it appears in the pattern, but how would we actually prove it?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1So 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 \[k1<\frac{(k+1)!}{1!+2!+...+k!}<k\] 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...

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1We know that (k+2)!=(k+1)! * (k+2) But I don't know what I'm gonna do what that bottom yet...

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I wonder if it would be easier to look at: (k1)(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
 one year ago
Best ResponseYou've already chosen the best response.1So (k+2)!=(k+1)!*(k+2) And we know what (k+1)! is in between

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1So 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)?

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0I don't know if this would help, something else I found on the problem. Like a start to a proof.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0Keep losing connect, sorry _

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0Running to class now, I'll be back around 2.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1\[k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)!\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1\[=(k1)(1!+...+k!)+k(k+1)! +2(1!+...+k!) \] \[\le (k+1)!+k(k+1)!+2(1!+...+k!) \] \[=(k+1)!(k+1)+2(1!+...+k!)\] \[\le (k+1)!(k+1)+(k+1)!=(k+1)!(k+1+1)=(k+2)!\] This should help... Now maybe you can prove the other side.

bbkzr31
 one year ago
Best ResponseYou've already chosen the best response.0Awesome, 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(2,2^161)) which is exactly what you had gotten.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(k(1!+...+k!+(k+1)!)=k(1!+...+k!)+k(k+1)! \) \(=(k1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0im 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
 one year ago
Best ResponseYou've already chosen the best response.11(1!+...+k!)+2(1!+...+k!) = 1(1!+...+k!)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0sorry 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
 one year ago
Best ResponseYou've already chosen the best response.1I don't understant 1+2=1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0sorry 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 : \(=(k1)(1!+...+k!)+k(k+1)! +\color{red}{2}(1!+...+k!) \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0i thought step2 should be : step2 : \(=(k1)(1!+...+k!)+k(k+1)! +\color{red}{1}(1!+...+k!)\) that 2 in ur step2 is throwing me off
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