2021 Fall AMC 12B Problems/Problem 1
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- The following problem is from both the 2021 Fall AMC 10B #1 and 2021 Fall AMC 12B #1, so both problems redirect to this page.
Problem
What is the value of 
Solution 1
We see that 
and 
each appear in the ones, tens, hundreds, and thousands digit exactly once. Since 
, we find that the sum is equal to ![\[10\cdot(1+10+100+1000)=\boxed{(\textbf{E})11,110}.\]](http://latex.artofproblemsolving.com/1/3/7/137d953cbcd2b520d1a0ccf3a8e5dfebcdf71f8f.png)
Note: it is equally valid to manually add all 4 numbers together to get the answer.
~kingofpineapplz
Solution 2
We have
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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. 
