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#2 2011-06-05 22:42:52
#4 2011-06-05 22:43:52
- veggieman001
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Re: Lets see who knows this kind of math...
five factorial
120
1*2*3*4*5
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#5 2011-06-05 22:44:00
Re: Lets see who knows this kind of math...
5! = 120
Not that hard. Let's see now, how about this? (Simplify; no calculators now! And yes, the answer is a whole number.)
(5log(125))/(log(2.5) + log(2))
Last edited by Harakou (2011-06-05 22:44:20)
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#6 2011-06-05 22:44:18
- soupoftomato
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Re: Lets see who knows this kind of math...
Above.
I'm glad to think that the community will always be kind and helpful, the language will always be a fun and easy way to be introduced into programming, the motto will always be: Imagine, Program, Share - Nomolos
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#7 2011-06-05 22:44:24
- Scratchthatguys
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Re: Lets see who knows this kind of math...
Hmm. Nope, wrong for me...
But 6! equals 3...
Anyways, it's 120. I read my sister's math book yesterday.
And before that I thought it was 240.
Last edited by Scratchthatguys (2011-06-05 22:46:09)
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#9 2011-06-05 22:44:54
- veggieman001
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Re: Lets see who knows this kind of math...
Scratchthatguys wrote:
Hmm. Nope, wrong...
But 6! equals 3...
Anyways, it's 120. I read my sister's math book yesterday.
6! isn't three
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#10 2011-06-05 22:45:29
- scratch_yoshi
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Re: Lets see who knows this kind of math...
5! = 1*2*3*4*5
1*2=2*3=6*4=24*5=120
6! = 1*2*3*4*5*6
120
* 6
720
Last edited by scratch_yoshi (2011-06-05 22:46:39)
Retired from Scratch, but just the language itself. Not the forums.
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#12 2011-06-05 22:47:06
#13 2011-06-05 22:47:49
Re: Lets see who knows this kind of math...
Harakou wrote:
5! = 120
Not that hard. Let's see now, how about this? (Simplify; no calculators now! And yes, the answer is a whole number.)
(5log(125))/(log(2.5) + log(2))
I have no clue, whats log?
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#18 2011-06-05 22:55:31
- scratch_yoshi
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Re: Lets see who knows this kind of math...
Wouldn't that be infinite? Or would it go up?
Retired from Scratch, but just the language itself. Not the forums.
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#19 2011-06-05 22:56:10
- technoguyx
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Re: Lets see who knows this kind of math...
JJROCKER wrote:
Just wondering... is it possible to have a negative factorial? Like -3! ?
http://mathforum.org/library/drmath/view/60851.html
Nope, according to that a negative factorial would have an undefined value due to division by zero.
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#22 2011-06-05 22:58:57
- scratch_yoshi
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Re: Lets see who knows this kind of math...
Actually, if factorals included 0, they'd all equal zero, just pointing that out 
Retired from Scratch, but just the language itself. Not the forums.
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#23 2011-06-05 22:59:12
Re: Lets see who knows this kind of math...
JJROCKER wrote:
Harakou wrote:
5! = 120
Not that hard. Let's see now, how about this? (Simplify; no calculators now! And yes, the answer is a whole number.)
(5log(125))/(log(2.5) + log(2))I have no clue, whats log?
log(x) is a logarithm of x with base 10. A logarithm is the inverse of a exponent. For a logarithm of x with base b and result a, b^a = x. So for example, log(100) = 2.
Of course, now that I posted that, I realized that it wasn't exactly the expression I intended. A better one that should be relatively easy to solve assuming you know the properties of logarithms would be this:
(125log(5))/(log(2.5) + log(2))
The basic properties are as follows:
log(a) + log(b) = log(ab)
log(a) - log (b) = log(a/b)
a * log(x) = log(x^a)
log (base b) of x = log(x)/log(b) (It's hard to express the base here because I don't have access to subscripts.)
Of course, I kinda feel like I'm intruding on your thread here. Not sure if you meant this as a specific discussion about factorials or a place to talk about mathematics in general and quiz each other. 
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