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−  {{duplicate[[2021 AMC 12B Problems2021 AMC 12B #15]] and [[2021 AMC 10B Problems2021 AMC 10B #20]]}}
 +  Please do not write problems in 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>
 
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−  ==Solution==
 
−  The prime factorization of <math>2021</math> is <math>43 \cdot 47</math>.
 
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−  ==See also==
 
−  {{AMC10 boxyear=2021ab=Bnumb=19numa=21}}
 
−  {{AMC12 boxyear=2021ab=Bnumb=14numa=16}}
 
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−  {{MAA Notice}}
 
Revision as of 16:39, 2 November 2020
Please do not write problems in before the contest has occurred.