AMC 10 Step by Step
2013 AMC 10BProblem 20P16-20~6 minPrint

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 a1a2ama_1\ge a_2\ge\cdots\ge a_m and b1b2bnb_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 a1b1|a_1-b_1| ?

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.