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Problem 900

AMC 10/12, early questions
Combinatorics Difficulty 4.0 Find the answer HMMT November · United States · 2023

Compute the smallest positive integer kk such that 4949 divides (2kk)\binom{2k}{k}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Solution:

The largest aa such that 7a7^{a} divides (2kk)\binom{2k}{k} is equal to the number of carries when you add k+kk + k in base 77, by Kummer's Theorem. Thus, we need two carries, so 2k2k must have at least 33 digits in base 77. Hence, 2k≥492k \geq 49, so k≥25k \geq 25. We know k=25k = 25 works because 25+25=347+347=101725 + 25 = 34_{7} + 34_{7} = 101_{7} has two carries.

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