Difference between revisions of "2006 AMC 12A Problems/Problem 24"
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== See also == | == See also == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
| + | by the multi-nomial theorem, the expressions of the two are: | ||
| + | |||
| + | <math>\sum{\frac{2006!}{a!b!c!}x^ay^bz^c}</math> | ||
| + | |||
| + | and: | ||
| + | |||
| + | <math>\sum{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c}</math> | ||
| + | |||
| + | respectively, where <math>a+b+c = 2006</math>. 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: | ||
| + | |||
| + | <math>{2006+2\choose 2} = 2015028</math> | ||
| + | |||
| + | 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: | ||
| + | |||
| + | if the exponent of <math>y</math> is 1, the exponent of <math>z</math> can be all even integers up to 2004, so 1003 terms. | ||
| + | |||
| + | if the exponent of <math>y</math> is 3, the exponent of <math>z</math> can go up to 2002, so 1002 terms. | ||
| + | |||
| + | <math>\vdots</math> | ||
| + | |||
| + | if the exponent of <math>y</math> is 2005$, then <math>z</math> can only be 0. So 1 term. | ||
| + | |||
| + | add them up we get <math>\frac{1003*1004}{2}</math> terms. However, we can switch the exponents of <math>y</math> and <math>z</math> and these terms will still have a negative sign. So there are a total of <math>1003*1004</math> negative terms. | ||
| + | |||
| + | Subtract this number from 2015028 we obtain <math>D. 1008016</math> as our answer. | ||
Revision as of 12:06, 12 July 2006
Problem
The expression
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
Solution
by the multi-nomial theorem, the expressions of the two are:
$\sum{\frac{2006!}{a!b!c!}x^ay^bz^c}$
and:
$\sum{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c}$
respectively, where $a+b+c = 2006$. 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:
$\binom{2006+2}{2} = 2015028$
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:
if the exponent of $y$ is 1, the exponent of $z$ can be all even integers up to $2004$, so 1003 solutions.
if the exponent of $y$ is 3, the exponent of $z$ can go up to $2002$, so 1002 solutions.
$\vdots$
if the exponent of $y$ is 2005$, then $z$ can only be $0$. So 1 solution.
add them up we get $\frac{1003*1004}{2}$ solutions. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003*1004$ negative terms.
Subtract this number from 2015028 we obtain $D. 1008016$ as our answer.
See also
by the multi-nomial theorem, the expressions of the two are:
and:
respectively, where 
. 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:
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.
