2021 Fall AMC 12B Problems/Problem 1
- 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.
Contents
[hide]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 that it is equally valid to manually add all four numbers together to get the answer.
~kingofpineapplz
Solution 2
We have ~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
Solution 4 (Brute Force)
We can simply add the numbers. .
Note: Although this would not take terribly long, it is not recommended to do this in a real contest.
~ cxsmi
Video Solution by Interstigation
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
For AMC 10: https://youtu.be/lC7naDZ1Eu4
For AMC 12: https://youtu.be/yaE5aAmeesc
~IceMatrix
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.