2021 Fall AMC 12B Problems/Problem 1: Difference between revisions
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== 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 | |
These problems are copyrighted © by the Mathematical Association of America.