Difference between revisions of "2006 AMC 12A Problems/Problem 24"
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== Solution == | == Solution == | ||
| − | By the [[Multinomial Theorem]], the | + | By the [[Multinomial Theorem]], the summands can be written as |
| − | < | + | <cmath>\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}</cmath> |
| − | and | + | and |
| − | < | + | <cmath>\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},</cmath> |
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are: | respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are: | ||
| − | < | + | <cmath>{2006+2\choose 2} = 2015028</cmath> |
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of <math>y</math> and <math>z</math> must be opposite. Now we find a pattern: | terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of <math>y</math> and <math>z</math> must be opposite. Now we find a pattern: | ||
Revision as of 19:05, 29 April 2008
Problem
The expression
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
Solution
By the Multinomial Theorem, the summands can be written as
![\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}\]](http://latex.artofproblemsolving.com/7/b/7/7b78526602fd2bf277218a45fd8d1fdcc3efcdf4.png)
and
![\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},\]](http://latex.artofproblemsolving.com/1/3/5/1353e7455d59a32ac3de47ace211b252a9bd0cf6.png)
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:
![\[{2006+2\choose 2} = 2015028\]](http://latex.artofproblemsolving.com/8/d/3/8d3ff6f9a5045933c1619fcab5e8bd4a48ee0457.png)
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of 
and 
must be opposite. Now we find a pattern:
if the exponent of is 1, the exponent of can be all even integers up to 2004, so 1003 terms.
if the exponent of is 3, the exponent of can go up to 2002, so 1002 terms.
if the exponent of is 2005, then can only be 0. So 1 term.
add them up we get 
terms. However, we can switch the exponents of and and these terms will still have a negative sign. So there are a total of 
negative terms.
Subtract this number from 2015028 we obtain 
as our answer.
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
