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

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{{duplicate|[[2021 AMC 12B Problems|2021 AMC 12B #15]] and [[2021 AMC 10B Problems|2021 AMC 10B #20]]}}
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==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>
 
Stores Giggles g yvnygnygbuhbhbhbihbuynfcutbddtff b8jhbihb🤣☺️🙃🤣
 
 
 
==Solution==
 
The prime factorization of <math>2021</math> is <math>43 \cdot 47</math>.
 
 
 
==See also==
 
{{AMC10 box|year=2021|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2021|ab=B|num-b=14|num-a=16}}
 
 
 
{{MAA Notice}}
 

Latest revision as of 15:40, 2 November 2020

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