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

(Created page with "{{duplicate|2021 AMC 12B #15 and 2021 AMC 10B #20}} ==Problem== The number <math>2021</math> is expressed in the form <cmat...")
 
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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>?
 
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>
 
<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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==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}}

Revision as of 14:25, 29 September 2020

The following problem is from both the 2021 AMC 12B #15 and 2021 AMC 10B #20, so both problems redirect to this page.

Problem

The number $2021$ is expressed in the form \[2021=\frac{a_1!a_2!...a_m!}{b_1!b_2!...b_n!},\] where $a_1 \geq a_2 \geq \cdots \geq a_m$ and $b_1 \geq b_2 \geq \cdots \geq b_n$ are positive integers and $a_1+b_1$ is as small as possible. What is $|a_1 - b_1|$? $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

See also

2021 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions