Difference between revisions of "2021 AMC 12B Problems/Problem 15"

Revision as of 16:40, 2 November 2020

Please do not write here before the contest has occurred.

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-{{duplicate|[[2021 AMC 12B Problems|2021 AMC 12B #15]] and [[2021 AMC 10B Problems|2021 AMC 10B #20]]}}
+Please do not write here before the contest has occurred.
-==Problem==
-The number <math>2021</math> is expressed in the form <cmath>2021=\frac{a_1!a_2!...a_m!}{b_1!b_2!...b_n!},</cmath> where <math>a_1 \geq a_2 \geq \cdots \geq a_m</math> and <math>b_1 \geq b_2 \geq \cdots \geq b_n</math> are positive integers and <math>a_1+b_1</math> is as small as possible. What is <math>|a_1 - b_1|</math>?
-<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math>