2021 Fall AMC 12A Problems/Problem 12

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)

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

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2021 Fall AMC 12A Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See Also