Difference between revisions of "2021 AMC 12B Problems/Problem 15"
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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> | ||
| + | |||
| + | ==Solution== | ||
| + | The prime factorization of <math>2021</math> is <math>43 \cdot 47</math>. | ||
==See also== | ==See also== | ||
Revision as of 13:36, 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 
is expressed in the form ![\[2021=\frac{a_1!a_2!...a_m!}{b_1!b_2!...b_n!},\]](http://latex.artofproblemsolving.com/1/c/e/1cecfa127f7ba50c3f1ba15366c47072997fff08.png)
where 
and 
are positive integers and 
is as small as possible. What is 
?
Solution
The prime factorization of is 
.
See also
| 2021 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
