Difference between revisions of "2021 Fall AMC 12A Problems/Problem 12"
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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> | ||
| − | 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 | + | 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(0), 6(1), 6(2), \ldots, 6(166).</cmath> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 18:49, 23 November 2021
Problem
What is the number of terms with rational coefficients among the 
terms in the expansion of ![$\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?$](http://latex.artofproblemsolving.com/8/7/8/8786632e043433ca2eb557b4dff52f3f1a486b1e.png)
Solution
By the Binomial Theorem, each term in the expansion is of the form ![\[\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},\]](http://latex.artofproblemsolving.com/3/0/2/302935a9eb411b60ee86db5546affe8540fffeb2.png)
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 
![\[6(0), 6(1), 6(2), \ldots, 6(166).\]](http://latex.artofproblemsolving.com/f/d/f/fdfa91aefec9f7df0db095b1df954cd312d72a58.png)
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
