Difference between revisions of "2021 Fall AMC 12A Problems/Problem 12"
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==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> | |
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+ | This problem is equivalent to counting the values of <math>k</math> such that both <math>\frac k3</math> and <math>\frac{1000-k}{2}</math> are integers. Note that <math>k</math> must be a multiple of <math>3</math> and a multiple of <math>2,</math> so <math>k</math> must be a multiple of <math>6.</math> There are <math>\boxed{\textbf{(C)}\ 167}</math> such values of <math>k:</math> <cmath>6\cdot0, 6\cdot1, 6\cdot2, \ldots, 6\cdot166.</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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==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 19:48, 23 November 2021
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
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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