Difference between revisions of "2021 Fall AMC 12B Problems/Problem 1"
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~kingofpineapplz | ~kingofpineapplz | ||
+ | |||
+ | == Solution 2 == | ||
+ | We have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 1234 + 2341 + 3412 + 4123 | ||
+ | & = 1111 \left( 1 + 2 + 3 + 4 \right) \\ | ||
+ | & = 11110 . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(E) }11,110}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/p9_RH4s-kBA | https://youtu.be/p9_RH4s-kBA |
Revision as of 21:13, 25 November 2021
- 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
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.