[KYA]

341

I might do the "other" solution. An OR of 1 and 15 should give the same result. That's originally why I calculated out 15 in binary.

[gabehabe]

341 is wrong.

I wrote that in C++ and I got a different result. [Told you, you can write your own in the language to test it, just not to find the answer of someone else's]

HINT: The answer is on that sheet. Somewhere. I think you've really overcomplicated it.

[KYA]

UGH. I missed a parentheses!

How did you shift a mixed number? in your head or on paper?

30/4 = 7.5

1 << 7.5?

This post has been edited by KYA: 22 December 2008 - 04:25 PM

[gabehabe]

Look at it more closely.

((10*3)/4)) * 2)

[KYA]

No. You mistyped () then. Look at the OP. (((1 << ((10*3)/4)) * 2) - (32*2)) / 192

() Takes precedent other then socpe resolution

I think writing a program is cheating. You should have to do it in your head/on paper like the OP says.

Anyways:

(((1 << ((10*3)/4)) * 2) - (32*2)) / 192
1<< 7.5 is really 1 << 7, so 128 * 2 = 256
-64 = 192
192/192
=1 <- Answer

This post has been edited by KYA: 22 December 2008 - 04:32 PM

[gabehabe]

You got it.

I apologise for getting it wrong in the first place. I'm tired.

That's my excuse and I'm sticking to it.

Essentially, I was right. I just read my own problem incorrectly.

I suck.

[KYA]

Further proof since we are talking about coding it:



Even not using integers it won't do half a bit

[KYA]

OK. Next math problem:

File's attached. Paint was being retarded. It's a series.

Attached File(s)

•  dic_math_problem__series_.doc (17K)
  Number of downloads: 169

[GWatt]

KYA, I can only see part of that equation.
the Y(n from 0 to 5) = (n + 1^n)/n!
is that right?

[KYA]

Hmmm, I used Mathtype in Word.

Anyway:

E (n = 0 to 5)

((n+1)^n)/n!

This post has been edited by KYA: 22 December 2008 - 06:39 PM

[GWatt]

OK, it shows up correctly in pages, but not openoffice

I forgot anything with series about 2 years ago, so this is the answer I got:
14041/120

This post has been edited by GWatt: 22 December 2008 - 07:01 PM

[Locke]

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...

This post has been edited by Locke37: 22 December 2008 - 07:50 PM

[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