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2013 AMC 10B Problem 20

Problem 20 of 25HarderNumber TheoryCounting & Probability

The number 20132013 is expressed in the form 2013=a1!a2!⋯am!b1!b2!⋯bn!,2013 = \frac {a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, where a1≥a2≥⋯≥ama_1 \ge a_2 \ge \cdots \ge a_m and b1≥b2≥⋯≥bnb_1 \ge b_2 \ge \cdots \ge b_n are positive integers and a1+b1a_1 + b_1 is as small as possible. What is ∣a1−b1∣?|a_1 - b_1|?

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Solution

The prime factorization is 2013=3⋅11⋅612013=3\cdot11\cdot61, so the numerator must contain a factor of 6161. Hence a1≥61a_1\ge61. But 61!61! also contains the prime factor 5959, which is not in 20132013, so the denominator must contain a factor of 5959. Hence b1≥59b_1\ge59. The lower bound a1+b1≥120a_1+b_1\ge120 is attainable because 2013=61! 11! 3!59! 10! 5!2013=\frac{61!\,11!\,3!}{59!\,10!\,5!}. Thus ∣a1−b1∣=61−59=2|a_1-b_1|=61-59=2, and the correct answer is B .

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Concepts: factorial · prime factorization

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