2021 AMC 12A Fall Problem 12

Below is the professionally curated solution for Problem 12 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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Concepts:binomial theoremdivisibility

Difficulty rating: 1630

12.

What is the number of terms with rational coefficients among the 10011001 terms in the expansion of (x23+y3)1000? \left(x\sqrt[3]{2} + y\sqrt{3}\right)^{1000}?

00

166166

167167

500500

501501

Solution:

The general term is (1000k)(x23)1000k(y3)k,\binom{1000}{k}(x\sqrt[3]{2})^{1000-k}(y\sqrt{3})^{k}, whose coefficient contains 2(1000k)/32^{(1000-k)/3} and 3k/2.3^{k/2}. This is rational exactly when 3(1000k)3 \mid (1000 - k) and kk is even.

Since 10001(mod3),1000 \equiv 1 \pmod 3, we need k1(mod3)k \equiv 1 \pmod 3 and kk even, which combine to k4(mod6).k \equiv 4 \pmod 6. The valid values k=4,10,,1000k = 4, 10, \ldots, 1000 number 100046+1=167.\dfrac{1000 - 4}{6} + 1 = 167.

Thus, the correct answer is C.

Problem 12 in Other Years