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07312007  #21 
Join Date: Oct 2006
Posts: 822

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; 07312007 at 04:08 PM.. 
08012007  #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. 
08012007  #23 
Join Date: Oct 2006
Posts: 822

Well, 29 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; 08012007 at 08:52 PM.. 
08012007  #24 
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Um, what's the "c" doing there? Are you sure you typed the LaTeX correctly?

08012007  #25 
Join Date: Oct 2006
Posts: 822

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; 08022007 at 05:35 PM.. 
08042007  #26 
Join Date: Oct 2006
Posts: 822

I'm gonna post the solution by 6:00 if no one answers by that time...
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08042007  #27 
Join Date: Oct 2006
Posts: 822

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: 2nd: Show more:
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. Last edited by Sillysidley; 08042007 at 05:06 PM.. 
08092007  #28 
Join Date: Oct 2006
Posts: 822

Anyone with a new problem?
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08092007  #29 
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Classic problem:
The coefficients of and in the expansion of are equal. Find n. 
08102007  #30 
Join Date: Oct 2006
Posts: 822

We have nC5=nC15, so n=20
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