2021 AMC 12A Fall Problem 12

Attempt Problem 12 of the 2021 AMC 12A Fall below, then check your answer against the professionally curated solution 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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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
Answer: C
Concepts:binomial theoremdivisibility
Small hint:
The term for index kk carries 21000−k3 3k2;2^{\frac{1000-k}{3}}\,3^{\frac{k}{2}}; rationality needs 1000−k1000 - k to be a multiple of 33 and kk to be a multiple of 22
Big hint:
These force k≡1(mod3)k \equiv 1 \pmod 3 and kk even, i.e. k≡4(mod6)k \equiv 4 \pmod 6
Solution:
The general term is (1000k)(x23)1000−k(y3)k,\binom{1000}{k}(x\sqrt[3]{2})^{1000-k}(y\sqrt{3})^{k}, whose coefficient contains 21000−k32^{\frac{1000-k}{3}} and 3k2.3^{\frac{k}{2}}. This is rational exactly when (1000−k)(1000 - k) is a multiple of 33 and kk is even.
Since 1000≡1(mod3),1000 \equiv 1 \pmod 3, we need k≡1(mod3)k \equiv 1 \pmod 3 and kk even, which combine to k≡4(mod6).k \equiv 4 \pmod 6. The valid values k=4,10,…,1000k = 4, 10, \ldots, 1000 number 1000−46+1=167.\dfrac{1000 - 4}{6} + 1 = 167.
Thus, the correct answer is C.
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