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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 12"

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<math>\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501</math>
 
<math>\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501</math>
  
−
==Solution==
+
==Solution 1==
  
 
By the Binomial Theorem, each term in the expansion is of the form <cmath>\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},</cmath> where <math>k\in\{0,1,2,\ldots,1000\}.</math>
 
By the Binomial Theorem, each term in the expansion is of the form <cmath>\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},</cmath> where <math>k\in\{0,1,2,\ldots,1000\}.</math>
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~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
== Solution 2 ==
 +
We have
 +
<cmath>
 +
\begin{align*}
 +
\left( x \sqrt[3]{2} + y \sqrt{3} \right)^{1000}
 +
& = \sum_{n = 0}^{1000} \binom{1000}{n} \left( x \sqrt[3]{2} \right)^n \left( y \sqrt{3} \right)^{1000-n} \\
 +
& = \sum_{n = 0}^{1000} \binom{1000}{n}
 +
2^{n/3} 3^{(1000-n)/2}
 +
x^n y^{1000-n} .
 +
\end{align*}
 +
</cmath>
 +
 +
Hence, the <math>n</math>th term is rational if and only if <math>2^{n/3} 3^{(1000-n)/2}</math> is rational.
 +
This holds if and only if <math>3 | n</math> and <math>2 | n</math>.
 +
These conditions are equivalent to <math>6 | n</math>.
 +
 +
Therefore, all rational terms are with <math>n = 0, 6, 6 \cdot 2, \cdots , 6 \cdot 166</math>.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(C) }167}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:43, 25 November 2021

Problem

What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?$

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501$

Solution 1

By the Binomial Theorem, each term in the expansion is of the form \[\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},\] where $k\in\{0,1,2,\ldots,1000\}.$

This problem is equivalent to counting the values of $k$ such that both $\frac k3$ and $\frac{1000-k}{2}$ are integers. Note that $k$ must be a multiple of $3$ and a multiple of $2,$ so $k$ must be a multiple of $6.$ There are $\boxed{\textbf{(C)}\ 167}$ such values of $k:$ \[6(0), 6(1), 6(2), \ldots, 6(166).\]

~MRENTHUSIASM

Solution 2

We have \begin{align*} \left( x \sqrt[3]{2} + y \sqrt{3} \right)^{1000} & = \sum_{n = 0}^{1000} \binom{1000}{n} \left( x \sqrt[3]{2} \right)^n \left( y \sqrt{3} \right)^{1000-n} \\ & = \sum_{n = 0}^{1000} \binom{1000}{n} 2^{n/3} 3^{(1000-n)/2} x^n y^{1000-n} . \end{align*}

Hence, the $n$th term is rational if and only if $2^{n/3} 3^{(1000-n)/2}$ is rational. This holds if and only if $3 | n$ and $2 | n$. These conditions are equivalent to $6 | n$.

Therefore, all rational terms are with $n = 0, 6, 6 \cdot 2, \cdots , 6 \cdot 166$.

Therefore, the answer is $\boxed{\textbf{(C) }167}$.

~Steven Chen (www.professorchenedu.com)

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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