2021 Fall AMC 12B Problems/Problem 1: Difference between revisions
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== Solution 1 == | == Solution 1 == | ||
We see that <math>1, 2, 3,</math> and <math>4</math> each appear in the ones, tens, hundreds, and thousands digit exactly once. Since <math>1+2+3+4=10</math>, we find that the sum is equal to <cmath>10\cdot(1+10+100+1000)=\boxed{(\textbf{E})11,110}.</cmath> | We see that <math>1, 2, 3,</math> and <math>4</math> each appear in the ones, tens, hundreds, and thousands digit exactly once. Since <math>1+2+3+4=10</math>, we find that the sum is equal to <cmath>10\cdot(1+10+100+1000)=\boxed{(\textbf{E })11,110}.</cmath> | ||
Revision as of 10:48, 1 January 2022
- 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/5/8/9/58997cf27d25b920b067e43c1473a862027763c4.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)
Solution 3
We see that the units digit must be 
, since 
is . But every digit from there, will be a 
since we have that each time afterwards, we must carry the 1 from the previous sum. The answer choice that satisfies these conditions is .
~~stjwyl
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
