2021 AMC 12B Problems/Problem 15

Revision as of 15:56, 19 October 2020 by Lnzhonglp (talk | contribs) (Problem)
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 $|a_1 - b_1|$? $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

The prime factorization of $2021$ is $43 \cdot 47$.

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

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