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Old 07-31-2007   #21
Sillysidley

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Quote:
Originally Posted by Archive View Post


Time to test the multinomial theorem...

What is ?
Let's try reviving this again.
It'd be a waste of time for me to actually do go through each exponent one by one, but here is the multinomial theorem:http://en.wikipedia.org/wiki/Multinomial_theorem
Try posting one with only x,y,z, much faster for peole to do.
REMEMBER, MIDDLE SCHOOL (or State Mathcount) LEVEL ONLY (seriously let's just do state mathcount level, good practice)

Find the greatest number that the sum of 3 consecutive odd numbers must be divisible by
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Last edited by Sillysidley; 07-31-2007 at 04:08 PM..
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Old 08-01-2007   #22
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It's 3, obviously.

Find the greatest number than cannot be expressed by sums of 2, 3, 4, 5, 6, 7, 8, and 9.
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Old 08-01-2007   #23
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Well, 2-9 can all be made (obviously) and you can get any any number higher by adding multiples of 10 (by adding 5s)
so 1.

Find

Oh, and if anyone cares, I meant to ask whats the largest number that must be a factor of any four consecutive odd numbers. (Three is too simple.)
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Last edited by Sillysidley; 08-01-2007 at 08:52 PM..
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Old 08-01-2007   #24
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Um, what's the "c" doing there? Are you sure you typed the LaTeX correctly?
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Old 08-01-2007   #25
Sillysidley

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Looks like I had to go to the next line for some reason..
Oh, I forgot to metion, this isone of my favorite mathcount problems. It has two different approaches (One with a formula, one with a kind of series.)
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Last edited by Sillysidley; 08-02-2007 at 05:35 PM..
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Old 08-04-2007   #26
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I'm gonna post the solution by 6:00 if no one answers by that time...
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Old 08-04-2007   #27
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Eh, close enough
The first one uses a telescoping series (basically where a bunch of things cancel out) and the second has a formula (and just to get better at proofs, I wanna prove it)
Show more:
We want all the terms except the first and last to cancel. So, we express it as:

2nd:
Show more:
There's a formula that says the sum of n numbers of this kind sum up to . In other words, We can then easily see the answer is .

Proof: Easy enough to prove by induction:
For n=1, the LHS=RHS=. Let's assume it holds for n. Then, for n+1;
And we're finally done
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Last edited by Sillysidley; 08-04-2007 at 05:06 PM..
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Old 08-09-2007   #28
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Anyone with a new problem?
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Old 08-09-2007   #29
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Classic problem:

The coefficients of and in the expansion of are equal. Find n.
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Old 08-10-2007   #30
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We have nC5=nC15, so n=20
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