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

m (Solution 1)
(Solution 2: Removed repetitive solution. Prof. Chen agreed to this through PM ...)
Line 12: Line 12:
  
 
~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 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

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png