Need help with a discrete math problem.

- anonymous

Need help with a discrete math problem.

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- anonymous

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- anonymous

@myininaya

- 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.

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- anonymous

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

- anonymous

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

- 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

Hmm, so would this be a proof by induction?

- anonymous

Induction should work.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

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

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- anonymous

sorry, lost connection for a couple minutes.

- anonymous

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

- myininaya

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

- anonymous

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

- 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

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

- 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

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

- 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

still thinking...

- anonymous

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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- anonymous

Keep losing connect, sorry -_-

- anonymous

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

- myininaya

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

- 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

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

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

- 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

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

- 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

I don't understant -1+2=1

- 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

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

yeah you are right

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