Difference between revisions of "2006 AMC 12A Problems/Problem 24"

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is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
 
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
  
<math> \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514</math><math>\mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028</math>
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<math> \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028</math>
 
 
  
 
== Solution ==
 
== Solution ==

Revision as of 19:22, 5 January 2008

Problem

The expression

$(x+y+z)^{2006}+(x-y-z)^{2006}$

is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

Solution

By the Multinomial Theorem, the expressions of the two are:

$\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}$

and:

$\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c}$

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:

${2006+2\choose 2} = 2015028$

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 terms.

if the exponent of $y$ is 3, the exponent of $z$ can go up to 2002, so 1002 terms.

$\vdots$

if the exponent of $y$ is 2005, then $z$ can only be 0. So 1 term.

add them up we get $\frac{1003\cdot1004}{2}$ terms. 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 $1003\cdot1004$ negative terms.

Subtract this number from 2015028 we obtain $D. 1008016$ as our answer.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
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