2021 Fall AMC 12B Problems/Problem 1: Difference between revisions
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
Revision as of 17:25, 16 August 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/c/d/9/cd92c0d1d365e60e535571a75b7976384857893e.png)
Note that it is equally valid to manually add all four numbers together to get the answer.
~kingofpineapplz
Solution 2
We have ![\[1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.\]](http://latex.artofproblemsolving.com/a/b/9/ab9f64db20cec332a367cb90f8cc5f2b812b5d36.png)
~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 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. 
