Difference between revisions of "2021 Fall AMC 12B Problems/Problem 1"
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~Education, the Study of Education | ~Education, the Study of Education | ||
+ | ==Video Solution by WhyMath== | ||
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+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{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}} | ||
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Revision as of 19:05, 26 September 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.
Contents
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 Education
Video Solution by WhyMath
~savannahsolver
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.