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== |
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 01:54, 26 November 2021
Contents
Problem
What is the number of terms with rational coefficients among the terms in the expansion of
Solution
By the Binomial Theorem, each term in the expansion is of the form where
This problem is equivalent to counting the values of such that both
and
are integers. Note that
must be a multiple of
and a multiple of
so
must be a multiple of
There are
such values of
~MRENTHUSIASM
Solution 2
We have
Hence, the th term is rational if and only if
is rational.
This holds if and only if
and
.
These conditions are equivalent to
.
Therefore, all rational terms are with .
Therefore, the answer is .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.