Fun beginner-level algebra riddle

[dxoigmn]

dxoigmn is not right i believe:

http://mathworld.wolfram.com/Parenthesis.html

[111787]

7! = 5040

5040/105 = 48

48^(1/X) = 2

X is not an integer.

(n-1)*(n-1)! = an integer

[dxoigmn]

7! = 5040

5040/105 = 48

48^(1/X) = 2

X is not an integer.

(n-1)*(n-1)! = an integer

You need to read that article above about how parenthesis work or consult any introductory algebra textbook. And perhaps brush up on order of operations/precedence rules.

(2n-1)!/(2^(n-1)*(n-1)!) for n=4

First we compute (2n-1)!:

(2n-1)!=(2*4-1)!=(8-1)!=7!=5040

Now we compute 2^(n-1):

2^(n-1)=2^(4-1)=2^3=8

Now we compute (n-1)!:

(4-1)!=3!=6

Now we compute 2^(n-1)*(n-1)! by replacement:

2^(n-1)*(n-1)!=8*6=48.

Now we compute (2n-1)!/(2^(n-1)*(n-1)!) by replacement:

(2n-1)!/(2^(n-1)*(n-1)!)=5040/48=105

So taking the first 4 odd integers and multiplying them together we get:

1 * 3 * 5 * 7 = 105

105=105

QED?

[111787]

So it is.

I read 2^(n-1)*(n-1)!
as
2^((n-1)*(n-1)!)
not
(2^(n-1))*(n-1)!

Thats thing about typing math, it doesn't always read quite right.

[dxoigmn]

Thats thing about typing math, it doesn't always read quite right.

Sure it does, just need practice.

[Darkness]

Hey, I just read this and decided to take a try at it.  Here's what I did, didnt wanna read anyone else's work cause it would ruin the suprise:
((2n-1)!)/((n-1)!*(2^(n-1)))

[Yoni]

Hey, I just read this and decided to take a try at it. Here's what I did, didnt wanna read anyone else's work cause it would ruin the suprise:
((2n-1)!)/((n-1)!*(2^(n-1)))

gj

[Dynobird]

Here's my solution. This may have been posted multiple times, I haven't checked because I'm too lazy, but oh well.

Problem: Put 1 * 3 * 5 * ... * (2n-1) into closed-form.

Answer: n! / ( 2^[n/2] * [n/2]! )         // this makes you not have to solve for (2n-1) what n would be. you can just plug n for the odd factorial.
given that [x=anything] is the greatest integer function and n can be any positive factorial
This will give you the "odd factorial" of it.

Work:
Finding a factorial of odd numbers is the same as finding the factorial of a number divided by its factorial of even numbers.
This way of saying may sound weird but here's an example: 7 * 5 * 3 * 1 = 8! / (8 * 6 * 4 * 2).
So why does this matter? We can solve for a "factorial of even numbers" now instead of a "factorial of odd numbers"
(I know the terminology probably doesn't exist). And solving for a factorial of even numbers is easy because we have a
good old fact on our side - every even number is divisible by two!

So lets start out with a simple case: 8 factorial for even numbers.
= 8 * 6 * 4 * 2
= 2(4) * 2(3) * 2(2) * 2(1)           // you're just pairing twos with all of the factors to make them even
= 2^4 * 4!

A general formula:
A factorial for even numbers, given an even factorial (like 8! instead of 7!), would be...
2^(n/2) * (n/2)! Where n is the factorial number.
So 10! for evens would be 2^(5) * 5!
Which comes out correct if you do it on your calculator.

So thus, given n as an even integer, which equals (the odd number that you want to get an "odd factorial" of) + 1,
the odd factorial is equal to: n! / [2^(n/2) * (n/2)!].

Yet, this is annoying, because it can confuse people as to what n is, whether it's the odd number or o + 1.
So a cool way to rewrite this is using the greatest integer function, which states that [x=anything] = the first integer before x so long as x isn't an integer itself.

so... [7.9] = 7.
[8] = 8.
[-7.9] = -8 

So we could rewrite this as n! / ( 2 ^ [n/2] * [n/2]! )
You see, if n is 7, then there are three even numbers in its factorial, and [3.5] = 3.

Interesting side note: I actually had to use the greatest integer function when programming the rational roots theorem for polynomials, since TI-83 BASIC doesn't have remainder function. I think I'll make a separate thread for "Problem: Program Rational Roots Theorem in TI-83 BASIC"
explaining that theorem =)