[rdsrds2120]

im not sure...because I just recently learned how to do this.... but somewhere around 108.9? Close?

[rdsrds2120]

Locke37, on 22 Dec, 2008 - 06:49 PM, said:

I don't believe you can calculate that...

It's been a while since I've dealt wit the rules of summations...but can you do 0!?

I don't recall...

~~It somehow works its way into equaling "1". Heres why:
If n! is defined as the product of all positive integers from 1 to n, then:
1! = 1*1 = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
...
n! = 1*2*3*...*(n-2)*(n-1)*n
and so on.
Logically, n! can also be expressed n*(n-1)!...

Therefore, at n=1, using n! = n*(n-1)!
1! = 1*0!
because anything multiplied by 0!=0!
it simplifies to
1=0!
and 0!=1
OK?

This post has been edited by rdsrds2120: 22 December 2008 - 08:38 PM

[KYA]

Yes the factorial of 0 is 1, i.e. 0! = 1


The correct answer is: 13081/120 or ~109.008

Gwatt the estimate of your answer is ~117

rdsrds2120 said 108.9, so he wins I guess

nobody got it exact

Let me work out the first two factorials:

1! = 1 * 0! = 1*1 = 1

0! = 1 //There's a proof of this, but I don't feel like typing it

This post has been edited by KYA: 22 December 2008 - 09:29 PM

[GWatt]

Quote

# The result must equate to an integer. [No decimals. At least not for now. Let's see how difficult this gets first!]

rules, KYA, rules .. .

[KYA]

Whoops.

But its a series not bitshifting or such. Throw me a bone here

This post has been edited by KYA: 22 December 2008 - 09:36 PM

[rdsrds2120]

I was doing good up until 4 or 5. then i had to start using paper. ha/ well, heres the next one then since I was closest to the last one. not nearly as hard, just uses some memory//(this one you can definitley use without pen/paper-- NO CHEATING!)



Find the value of the question mark.

This post has been edited by rdsrds2120: 22 December 2008 - 10:06 PM

[GWatt]

25

[rdsrds2120]

right, like I said, just some short-term memory.

[rdsrds2120]

well, to keep things going, these are pretty fun. You can do whatever you like to the following sets of four numbers to set up an equation as long as their outcomes equal positive 21, here are the sets .oh, and they all work, i have done them.
These are pretty much guess and check, so if you figure them out please have another problem ready. Thx

Ex: 1 1 1 7: 7 ( 1 + 1 + 1 ) = 7x3 = 21
3 4 4 7: 4 x 7 - 4 - 3 = 28 - 7 = 21

168 1 2 3 :
6 8 10 12:
11 9 5 7:
10 2 42 3:
50 4 2 1:
50 2 3 1:

This post has been edited by rdsrds2120: 22 December 2008 - 11:25 PM

[Gloin]

168 1 2 3 : 1 * 168 - 2^3 = 21
6 8 10 12: 6 + (12*10) / 8 = 21 or (12 + 10) - (8 / 6) = 21 (integer division)
11 9 5 7: (9 + 5 - 11) * 7 = 21
10 2 42 3: 42 / (10 - 2^3) = 21 or 42 / ((10 / 2) - 3) = 21
50 4 2 1: (50 / 2) - (4 * 1) = 21
50 2 3 1: (50 / 2) - (3+1) = 21

Sorry, but I'm too lazy to figure out anything better but since I did this without any aid once, calculate 6^6
Don't post unless you calculate it without aid (not even pen and paper)

This post has been edited by Gloin: 23 December 2008 - 11:23 AM

[rdsrds2120]

mmm... After a lot of thinking. It might be somewhere around 40, 608.

[Gloin]

That's a bit off and I expect the exact number.
The number certainly makes you think of 2^16 = 65536

This post has been edited by Gloin: 23 December 2008 - 12:31 PM

[rdsrds2120]

ech. give me a couple more minutes and i will have it
i know it has to be kinda close.

[Gloin]

You're more than 10% off.

[rdsrds2120]

ok! I got it now. 46,656. Thats some big multplication just for in my head. It was easier when instea of actually doing 6*6*6*6*6*6
i did 36*36*36. less room for error. Alright. The next one.
Over the 12 days of christmas, how many total presents did I recieve?(thought I could incorporate some holiday in with this one.
Oh, and btw, the question follows the song)

This post has been edited by rdsrds2120: 23 December 2008 - 01:37 PM