Question

How should I do this?

Question attached.

Answer 1

@.Sam. @whpalmer4

Answer 2

@TuringTest

Answer 3

what is (2n+1)!?

Answer 4

do you just delete tags as time goes by?

Answer 5

Factorial right?

Yeah, I do, so that it does not crowd the area; it helps OpenStudy. :)

Answer 6

@amistre64 ? :|

Answer 7

then your best bet is to start expanding it out, or at least that what I would do to start with, then you can see it for the polymonial that it is and divide

Answer 8

But

(2n+1)! = (2n+1)(2n)(2n-1)(2n-2)(2n-3)...
(n+1)\(^2\) = n\(^2\) + 2n + 1

How should I expand that completely?
And how should I exactly divide this?

Answer 9

@ranga
@ParthKohli
@mathslover
@mathstudent55

Answer 10

a! = a(a-1)(a-2)(a-3)...3.2.1

let a=2n+1

(2n+1)(2n)(2n-1)(2n-2)...3.2.1; let me consider my idea some more

Answer 11

and it has to be some (n+1)^2 = (2n+1)! right?

Answer 12

one poly is a factor of another if they have the same zeros comes to mind

Answer 13

No, (n+1)^2 = \(\lambda\)*(2n+1)!
Where \(\lambda\) is a natural number.

It must be divide it. :/

Answer 14

\[\frac{p(x)~q(x)~r(x)~...~k(x)}{p(x)}\]is one thought i have going, which makes me think that when n=-1 is the only solution; but then that is not in the domain so it has no solutions?

Answer 15

It does.

When n = 3

(n+1)^2 = 16

(2n+1)! = 5040

5040/16 = \(\lambda\) [Which is a natural number]

Answer 16

but that thought may not be valid ....

http://www.wolframalpha.com/input/?
i=table+%5B%282n%2B1%29%21%2F%28n%2B1%29%5E2%5D%2C+n%3D1..100

Answer 17

it looks like when (n+1)^2 is prime it fails

Answer 18

The only problem is, after this, when n increases, say 10
21 ! is very hard to calculate on a normal calculator.
So I can't even check for them.

Answer 19

can you think of a reason why (n+1)^2 is a prime that it would fail?

Answer 20

But (n+1)^2 is never prime.

Nope, I have no idea. I think I'll try graphing the function once and see what's happening.

Answer 21

ahh, in simplest form its a prime ....

Answer 22

like 3/5 = 6/10 but 10 is not prime

Answer 23

@zepdrix

Answer 24

I think this question is entirely based on Number theory?

Answer 25

(n+1)(n+1) simplifies to a prime if n+1 is a prime; which simply means that n+1 can be factored from the top

i think so too

Answer 26

We cannot get in (n+1) in the numerator can we?

Answer 27

i dont see why not ... i just dont see how yet

Answer 28

\[\frac{(2n+1)!}{p_1~p_2}\]
for example

11! = 11.10.9.8.7.6.5.4.3.2.1 has plenty of primes in it

Answer 29

but then n+1 = 6

let n=6

13.12.11.10.9.8.7.6.5.4.3.2.1
--------------------------
7 . 7

Answer 30

All I observed from graphing was that, the factorial function increases just too darn fast!

Answer 31

My only question is, how is it helping us? ?

Answer 32

Yeah, 14 is not there.

Answer 33

let n=some prime, minus 1

\[\frac{(2p+1).(2p).(2p-1).(2p-2)...3.2.1}{p.p}\]

Answer 34

But for 11. It works.

Answer 35

ugh, the top is p-1, not p

Answer 36

\[\frac{(2(p-1)+1).(2(p-1).(2(p-1)-1).(2(p-1)-2)...3.2.1}{p.p}\]
\[\frac{(2p-1).(2p-2).(2p-3).(2p-4)...3.2.1}{p.p}\]

Answer 37

somewhere in the top, it reduces to only 1 p to factor out

Answer 38

2p-k = p

when k=p

Answer 39

Yep.
And that is when

2n-1,2n-2,2n-3 reaches 2n-n. which is n.

Answer 40

Yeah. lol

Answer 41

and 2p-k = p^2 when k = ??

Answer 42

Why?

one p is already lost, with the 2n - 0 = 2n = 2p

Answer 43

hmm, i was trying to consider if there was another factor of p, if we can get rid of both ps then ...

Answer 44

This question is annoying me now.

Answer 45

(2p-k)(2p-h) = p^2

is this possible?

Answer 46

Only, if h=k=p

Answer 47

\[4p^2-2(h+k)p+hk=p^2\]
\[3p^2-2(h+k)p+hk=0\]
but k=p is one of them
\[3p^2-2(h+p)p+hp=0\]
\[3p^2-2hp-2p^2+hp=0\]
\[p^2-2hp+hp=0\]
\[p^2-hp=0~:~h=p=k\]

Answer 48

you know if your just going to not work it all out in pretty latex code im just gonna leave lol

Answer 49

can we work this for evens and odds and come to some by case conslusions?

Answer 50

It's just that, I have to submit 3-4 question like this, in 2 hours, and I have no idea how to do this, it is a little bit tiring/annoying to sit doing one question for hours. :|

Answer 51

yeah, which is why mtDew helps ;)

Answer 52

if it wasnt for the wolfs table and seeing the primes as denominators, id have no clue

Answer 53

hehe, I'll try that next time.

Answer 54

I just noticed something!

When we have a prime number - 1 set as n.

Then the (2n+1)! never really reaches a multiple of (n+1) a second time.

Answer 55

@amistre64

I think we can conclude, NO
n = prime number - 1
can satisfy the required. :D

Answer 56

Oh wait. I think that is for all the 'n'.

But for all the composite numbers,

n \(\neq\) prime - 1

Then it IS always divisible!

Answer 57

I think the answer simply is:

The no. of composite numbers [less than 100] - the no. of prime numbers [less than 100].
= 75

Answer 58

@ganeshie8 ?

Answer 59

\[\dfrac{(2n+1)!}{(n+1)^2} = \dfrac{1.2.3 \cdots (2n+1)}{(n+1)^2}\]

If \(n+1\) is composite then :
there will exist \(n+1 = m\times n\) such that \(m,n < \dfrac{n+1}{2}\)
To conclude the proof, just notice that \( \dfrac{n+1}{2} \lt \dfrac{n+1}{2}\times 2 \lt 2n+1\)

you still need to prove primes wont work.

Answer 60

let n=(2e-1)

\[\frac{(2(2e-1)+1).(2(2e-1).(2(2e-1)-1).(2(2e-1)-2)...3.2.1}{2e.2e}\]
\[\frac{(4e-1).(4e-2).(4e-3).(4e-4)...3.2.1}{2e.2e}\]
4e-k = 2e, when k=2e

(4e-k)(4e-h) = 4e^2

(2e)(4e-h) = 4e^2

4e-h = 2e, h = 2(2e) = 2k which is easily doable

since k=2,4,6,8,...,50 from the set of 1,2,3,...,199

Answer 61

odds most likely work the same

Answer 62

@ganeshie8

Is (2n-1)! = (2n-1)(2n-2)(2n-3)......(2n-n)(n-1)(n-2)(n-3).....3.2.1 ?

Answer 63

If so, I can prove, primes won't work.

Answer 64

@ganeshie8

What doe we understand from:

(n+1)/2 < (n+1) < (2n+1)

And

If a.b = n+1
a,b <= (n+1)/2 right?

Answer 65

yes here is the full work for both prime and composite :
\(\dfrac{(2n+1)!}{(n+1)^2} = \dfrac{1.2.3 \cdots (2n+1)}{(n+1)^2} \)

\(\color{red}{\text{If n+1 is composite : }}\)
there will always exist \(n+1=m \times n\) such that \(m \le \frac{n+1}{2} \) and \(n \le \frac{n+1}{2} \)

next notice that \(\dfrac{n+1}{2} \lt \dfrac{n+1}{2}\times 2 \lt \dfrac{n+1}{2}\times 3 \lt 2n+1\)

That means for each of m, n there are atleast 3 multiples in \( (2n+1)! \)
since we need only two multiples of m for m^2 to divide (2n+1)! we are done.

\(\color{red}{\text{If n+1 is prime : }}\)
the two successive least positive multiples of \(n+1\) are : \(n+1\), \(2(n+1)\)
clearly \(2(n+1) = 2n+2 \) does not exist in \((2n+1)!\)
so there are no solutions when \(n+1\) is prime

Answer 66

I was doing the exact same thing!
:D

THANKS! :D

Answer 67

yw :)