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

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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution by TheBeautyofMath==
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https://youtu.be/ToiOlqWz3LY?t=169
  
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== Solution 2 ==
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~IceMatrix
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We have
 
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<cmath>
 
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\begin{align*}
 
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\left( x \sqrt[3]{2} + y \sqrt{3} \right)^{1000}
 
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& = \sum_{n = 0}^{1000} \binom{1000}{n} \left( x \sqrt[3]{2} \right)^n \left( y \sqrt{3} \right)^{1000-n} \\
 
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& = \sum_{n = 0}^{1000} \binom{1000}{n}
 
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2^{n/3} 3^{(1000-n)/2}
 
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x^n y^{1000-n} .
 
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\end{align*}
 
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</cmath>
 
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Hence, the <math>n</math>th term is rational if and only if <math>2^{n/3} 3^{(1000-n)/2}</math> is rational.
 
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This holds if and only if <math>3 | n</math> and <math>2 | n</math>.
 
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These conditions are equivalent to <math>6 | n</math>.
 
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Therefore, all rational terms are with <math>n = 0, 6, 6 \cdot 2, \cdots , 6 \cdot 166</math>.
 
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Therefore, the answer is <math>\boxed{\textbf{(C) }167}</math>.
 
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~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}}

Latest revision as of 21:01, 7 April 2022

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

Video Solution by TheBeautyofMath

https://youtu.be/ToiOlqWz3LY?t=169

~IceMatrix

See Also

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

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