Difference between revisions of "2021 Fall AMC 12B Problems/Problem 1"
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− | {{duplicate|[[2021 Fall AMC 10B Problems | + | {{duplicate|[[2021 Fall AMC 10B Problems/Problem 1|2021 Fall AMC 10B #1]] and [[2021 Fall AMC 12B Problems/Problem 1|2021 Fall AMC 12B #1]]}} |
+ | |||
==Problem== | ==Problem== | ||
− | What is the value of <math>1234+2341+3412+4123 | + | What is the value of <math>1234 + 2341 + 3412 + 4123</math> |
− | <math> | + | <math>\textbf{(A)}\: 10{,}000\qquad\textbf{(B)} \: 10{,}010\qquad\textbf{(C)} \: 10{,}110\qquad\textbf{(D)} \: 11{,}000\qquad\textbf{(E)} \: 11{,}110</math> |
== Solution 1 == | == Solution 1 == | ||
− | We see that <math>1, 2, 3,</math> and <math>4</math> each appear in the ones, tens, hundreds, and thousands digit exactly once. Since <math>1+2+3+4=10</math>, we find that the sum is equal to <cmath>10\cdot(1+10+100+1000)=\boxed{ | + | We see that <math>1, 2, 3,</math> and <math>4</math> each appear in the ones, tens, hundreds, and thousands digit exactly once. Since <math>1+2+3+4=10</math>, we find that the sum is equal to <cmath>10\cdot(1+10+100+1000)=\boxed{\textbf{(E)} \: 11{,}110}.</cmath> |
+ | Note that it is equally valid to manually add all four numbers together to get the answer. | ||
+ | |||
+ | ~kingofpineapplz | ||
+ | == Solution 2 == | ||
+ | We have <cmath>1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.</cmath> | ||
+ | ~Steven Chen (www.professorchenedu.com) | ||
− | + | == Solution 3== | |
+ | We see that the units digit must be <math>0</math>, since <math>4+3+2+1</math> is <math>0</math>. But every digit from there, will be a <math>1</math> since we have that each time afterwards, we must carry the <math>1</math> from the previous sum. The answer choice that satisfies these conditions is <math>\boxed{\textbf{(E)} \: 11{,}110}</math>. | ||
+ | ~stjwyl | ||
+ | |||
+ | == Solution 4 (Brute Force)== | ||
+ | We can simply add the numbers. <math>1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{\textbf{(E)}}</math>. | ||
+ | |||
+ | Note: Although this would not take terribly long, it is not recommended to do this in a real contest. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
− | |||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/p9_RH4s-kBA | https://youtu.be/p9_RH4s-kBA | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/I9lSM0hO39M | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/3UZHiV65WXU | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | For AMC 10: https://youtu.be/lC7naDZ1Eu4 | ||
+ | |||
+ | For AMC 12: https://youtu.be/yaE5aAmeesc | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Latest revision as of 00:48, 30 January 2024
- 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
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.