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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>
  
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==Solution 1==
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==Solution==
  
 
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>

Revision as of 00:54, 26 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

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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