2021 Fall AMC 12B Problems/Problem 1: Difference between revisions
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==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
For AMC 10: https://youtu.be/lC7naDZ1Eu4 | For AMC 10: https://youtu.be/lC7naDZ1Eu4 | ||
For AMC 12: https://youtu.be/yaE5aAmeesc | For AMC 12: https://youtu.be/yaE5aAmeesc | ||
~IceMatrix | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|before=First Problem|num-a=2}} | {{AMC10 box|year=2021 Fall|ab=B|before=First Problem|num-a=2}} | ||
{{AMC12 box|year=2021 Fall|ab=B|before=First Problem|num-a=2}} | {{AMC12 box|year=2021 Fall|ab=B|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:45, 29 December 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
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
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 | |
These problems are copyrighted © by the Mathematical Association of America.