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

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The expression  
 
The expression  
  
−
<math>(x+y+z)^{2006}+(x-y-z)^{2006}</math>
+
<cmath>(x+y+z)^{2006}+(x-y-z)^{2006}</cmath>
  
 
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>
+
<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 1==
−
by the multi-nomial theorem, the expressions of the two are:
+
By the [[Multinomial Theorem]], the summands can be written as
  
−
$\sum{\frac{2006!}{a!b!c!}x^ay^bz^c}$
+
<cmath>\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}</cmath>
  
−
and:
+
and
  
−
$\sum{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c}$
+
<cmath>\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},</cmath>
  
−
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:
+
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:
  
−
$\binom{2006+2}{2} = 2015028$
+
<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 $y$ and $z$ 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:
  
−
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 <math>y</math> is <math>1</math>, the exponent of <math>z</math> can be all even integers up to <math>2004</math>, so there are <math>1003</math> terms.
  
−
if the exponent of $y$ is 3, the exponent of $z$ can go up to $2002$, so 1002 solutions.
+
if the exponent of <math>y</math> is <math>3</math>, the exponent of <math>z</math> can go up to <math>2002</math>, so there are <math>1002</math> terms.
  
−
$\vdots$
+
<math>\vdots</math>
  
−
if the exponent of $y$ is 2005$, then $z$ can only be $0$.  So 1 solution.
+
if the exponent of <math>y</math> is <math>2005</math>, then <math>z</math> can only be 0, so there is <math>1</math> term.
  
−
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.
+
If we add them up, we get <math>\frac{1003\cdot1004}{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\cdot1004</math> negative terms.
  
−
Subtract this number from 2015028 we obtain $D. 1008016$ as our answer.
+
By subtracting this number from 2015028, we obtain <math>\boxed{\mathrm{D}}</math> or <math>1,008,016</math> as our answer.
 +
 
 +
==Solution 2==
 +
 
 +
Alternatively, we can use a [[generating function]] to solve this problem.
 +
The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of <math>k</math>). In other words, we want to find <math>f(x)</math> where the coefficient of <math>x^k</math> equals the number of unique terms in <math>(x+y+z)^k + (x-y-z)^k</math>.
 +
 
 +
 
 +
First, we note that all unique terms in the expression have the form, <math>Cx^ay^bz^c</math>, where <math>a+b+c=k</math> and <math>C</math> is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of <math>(x+y+z)^k + (x-y-z)^k</math> is
 +
<cmath>(1+x+x^2+x^3\cdots)^3 = \frac{1}{(1-x)^3}</cmath>
 +
 
 +
 
 +
Secondly, we note that a certain number of terms of the form, <math>Cx^ay^bz^c</math>, do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when <math>1 \equiv b+c\text{ (mod }2\text{)}</math> because every unique term is of the form:
 +
<cmath>\binom{k}{a,b,c}x^ay^bz^c+(-1)^{b+c}\binom{k}{a,b,c}x^ay^bz^c</cmath>
 +
for all possible <math>a,b,c</math>.
 +
 
 +
 
 +
Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of <math>k</math>. With some thought, we see that this desired generating function is the following:
 +
<cmath>2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}</cmath>
 +
 
 +
 
 +
Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in <math>(x+y+z)^k + (x-y-z)^k</math>, our initial goal:
 +
<cmath>\frac{1}{(1-x)^3}-\frac{2x}{(1-x)^3(1+x)^2} = \frac{x^2+1}{(1-x)^3(1+x)^2}</cmath>
 +
which equals
 +
<cmath>(x^2+1)(1+x+x^2\cdots)^3(1-x+x^2-x^3\cdots)^2</cmath>
 +
 
 +
 
 +
The coefficient of <math>x^{2006}</math> of the above expression equals
 +
<cmath>\sum_{a=0}^{2006}\binom{2+a}{2}\binom{1+2006-a}{1}(-1)^a + \sum_{a=0}^{2004}\binom{2+a}{2}\binom{1+2004-a}{1}(-1)^a</cmath>
 +
 
 +
 
 +
Evaluating the expression, we get <math>1008016</math>, as expected.
 +
 
 +
== Solution 3 ==
 +
 
 +
Define <math>P</math> such that <math>P=y+z</math>. Then the expression in the problem becomes:
 +
<math>(x+P)^{2006}+(x-P)^{2006}</math>.
 +
 
 +
Expanding this using binomial theorem gives <math>(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)</math>, where <math>n=2006</math> (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).
 +
 
 +
Simplifying gives: <math>2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)</math>. Note that two terms that come out of different powers of <math>P</math> cannot combine and simplify, as their exponent of <math>x</math> will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, <math>P^k=(y+z)^k</math> will have <math>k+1</math> terms, so the answer is <math>1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{D}</math>.
 +
 
 +
==Solution 4==
 +
 
 +
Using stars and bars we know that <math>(x+y+z)^{2006}</math> has <math>{2006+2\choose 2}
 +
</math> or <math>2015028</math> different terms. Now we need to find out how many of these terms are canceled out by <math>(x-y-z)^{2006}</math>. We know that for any term(let's say <math>x^{a}(-y)^{b}(-z)^{c}</math>) where <math>a+b+c=2006</math> of the expansion of  <math>(x-y-z)^{2006}</math> is only going to cancel out with the corresponding term <math>x^{a}y^{b}z^{c}</math> if only <math>b</math> is odd and <math>c</math> is even or <math>b</math> is even and <math>c</math> is odd. Now let's do some casework to see how many terms fit this criteria:
 +
 
 +
Case 1:    <math>a</math> is even 
 +
 
 +
Now we know that <math>a</math> is even and <math>a+b+c=2006</math>. Thus <math>b+c</math> is also even or both <math>b</math> and <math>c</math> are odd or both <math>b</math> and <math>c</math> are even. This case clearly fails the above criteria and there are 0 possible solutions.
 +
 
 +
Case 2:    <math>a</math> is odd
 +
 
 +
Now we know that <math>a</math> is odd and <math>a+b+c=2006</math>. Thus <math>b+c</math> is odd and <math>b</math> is odd and <math>c</math> is even or <math>b</math> is even and <math>c</math> is odd. All terms that have <math>a</math> being odd work.
 +
 
 +
We now need to figure out how many of the terms have <math>a</math> as a odd number. We know that <math>a</math> can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus <math>\frac{1003}{2007}</math> of the <math>2015028</math> terms will cancel out and  <math>\frac{1004}{2007}</math> of the terms will work. Thus there are <math>(\frac{1004}{2007})(2015028)</math> terms. This number comes out to be <math>1,008,016</math> <math>\implies \boxed{D}</math>
 +
(Author: David Camacho)
 +
 
 +
==Solution 5==
 +
Noticing how <math>y</math> and <math>z</math> are negative in the second part of the expression, let <math>x=a</math> and <math>y+z = b</math>. Then we get <cmath>(a+b)^{2006} + (a-b)^{2006}</cmath>
 +
We know that the terms that don't cancel out have even powers of <math>a</math> and <math>b</math>. Our expansion will be in the form of
 +
<cmath>2a^{2006} + x_1a^{2004}b^{2} + x_2a^{2002}b^{4} + \cdots + 2b^{2006}</cmath>
 +
Note that <math>b^n = (x+y)^n</math> has <math>n+1</math> terms. Furthermore, the current expression is irreducible as each term has a different <math>x</math> power. Thus, when we write <math>a</math> and <math>b</math> back to their original terms, we will have <math>1+3+5+ \cdots + 2007 = 1004^2 = 1,008,016 \implies \boxed{D}</math>
 +
 
 +
-smartguy888
  
 
== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
 +
 +
{{AMC12 box|year=2006|ab=A|num-b=23|num-a=25}}
 +
{{MAA Notice}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 16:37, 17 September 2023

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 1

By the Multinomial Theorem, the summands can be written as

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

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 there are $1003$ terms.

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

$\vdots$

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

If we 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.

By subtracting this number from 2015028, we obtain $\boxed{\mathrm{D}}$ or $1,008,016$ as our answer.

Solution 2

Alternatively, we can use a generating function to solve this problem. The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of $k$). In other words, we want to find $f(x)$ where the coefficient of $x^k$ equals the number of unique terms in $(x+y+z)^k + (x-y-z)^k$.


First, we note that all unique terms in the expression have the form, $Cx^ay^bz^c$, where $a+b+c=k$ and $C$ is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of $(x+y+z)^k + (x-y-z)^k$ is \[(1+x+x^2+x^3\cdots)^3 = \frac{1}{(1-x)^3}\]


Secondly, we note that a certain number of terms of the form, $Cx^ay^bz^c$, do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when $1 \equiv b+c\text{ (mod }2\text{)}$ because every unique term is of the form: \[\binom{k}{a,b,c}x^ay^bz^c+(-1)^{b+c}\binom{k}{a,b,c}x^ay^bz^c\] for all possible $a,b,c$.


Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of $k$. With some thought, we see that this desired generating function is the following: \[2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}\]


Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in $(x+y+z)^k + (x-y-z)^k$, our initial goal: \[\frac{1}{(1-x)^3}-\frac{2x}{(1-x)^3(1+x)^2} = \frac{x^2+1}{(1-x)^3(1+x)^2}\] which equals \[(x^2+1)(1+x+x^2\cdots)^3(1-x+x^2-x^3\cdots)^2\]


The coefficient of $x^{2006}$ of the above expression equals \[\sum_{a=0}^{2006}\binom{2+a}{2}\binom{1+2006-a}{1}(-1)^a + \sum_{a=0}^{2004}\binom{2+a}{2}\binom{1+2004-a}{1}(-1)^a\]


Evaluating the expression, we get $1008016$, as expected.

Solution 3

Define $P$ such that $P=y+z$. Then the expression in the problem becomes: $(x+P)^{2006}+(x-P)^{2006}$.

Expanding this using binomial theorem gives $(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)$, where $n=2006$ (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).

Simplifying gives: $2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)$. Note that two terms that come out of different powers of $P$ cannot combine and simplify, as their exponent of $x$ will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, $P^k=(y+z)^k$ will have $k+1$ terms, so the answer is $1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{D}$.

Solution 4

Using stars and bars we know that $(x+y+z)^{2006}$ has ${2006+2\choose 2}$ or $2015028$ different terms. Now we need to find out how many of these terms are canceled out by $(x-y-z)^{2006}$. We know that for any term(let's say $x^{a}(-y)^{b}(-z)^{c}$) where $a+b+c=2006$ of the expansion of $(x-y-z)^{2006}$ is only going to cancel out with the corresponding term $x^{a}y^{b}z^{c}$ if only $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. Now let's do some casework to see how many terms fit this criteria:

Case 1: $a$ is even

Now we know that $a$ is even and $a+b+c=2006$. Thus $b+c$ is also even or both $b$ and $c$ are odd or both $b$ and $c$ are even. This case clearly fails the above criteria and there are 0 possible solutions.

Case 2: $a$ is odd

Now we know that $a$ is odd and $a+b+c=2006$. Thus $b+c$ is odd and $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. All terms that have $a$ being odd work.

We now need to figure out how many of the terms have $a$ as a odd number. We know that $a$ can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus $\frac{1003}{2007}$ of the $2015028$ terms will cancel out and $\frac{1004}{2007}$ of the terms will work. Thus there are $(\frac{1004}{2007})(2015028)$ terms. This number comes out to be $1,008,016$ $\implies \boxed{D}$ (Author: David Camacho)

Solution 5

Noticing how $y$ and $z$ are negative in the second part of the expression, let $x=a$ and $y+z = b$. Then we get \[(a+b)^{2006} + (a-b)^{2006}\] We know that the terms that don't cancel out have even powers of $a$ and $b$. Our expansion will be in the form of \[2a^{2006} + x_1a^{2004}b^{2} + x_2a^{2002}b^{4} + \cdots + 2b^{2006}\] Note that $b^n = (x+y)^n$ has $n+1$ terms. Furthermore, the current expression is irreducible as each term has a different $x$ power. Thus, when we write $a$ and $b$ back to their original terms, we will have $1+3+5+ \cdots + 2007 = 1004^2 = 1,008,016 \implies \boxed{D}$

-smartguy888

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

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