Difference between revisions of "2021 Fall AMC 12B Problems/Problem 1"

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{{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]]}}
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{{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]]}}
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==Problem==
 
==Problem==
What is the value of <math>1234+2341+3412+4123?</math>
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What is the value of <math>1234 + 2341 + 3412 + 4123</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>
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<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{(\textbf{E})11,110}.</cmath>
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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>
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Note that it is equally valid to manually add all four numbers together to get the answer.
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~kingofpineapplz
  
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== Solution 2 ==
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We have <cmath>1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.</cmath>
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~Steven Chen (www.professorchenedu.com)
  
Note: it is equally valid to manually add all 4 numbers together to get the answer.
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== Solution 3==
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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>.
  
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~stjwyl
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== Solution 4 (Brute Force)==
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We can simply add the numbers. <math>1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{\textbf{(E)}}</math>.
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Note: Although this would not take terribly long, it is not recommended to do this in a real contest.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
~kingofpineapplz
 
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/p9_RH4s-kBA
 
https://youtu.be/p9_RH4s-kBA
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 +
==Video Solution==
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https://youtu.be/I9lSM0hO39M
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~Education, the Study of Everything
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==Video Solution by WhyMath==
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https://youtu.be/3UZHiV65WXU
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~savannahsolver
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/lC7naDZ1Eu4
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For AMC 12: https://youtu.be/yaE5aAmeesc
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~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.

Problem

What is the value of $1234 + 2341 + 3412 + 4123$

$\textbf{(A)}\: 10{,}000\qquad\textbf{(B)} \: 10{,}010\qquad\textbf{(C)} \: 10{,}110\qquad\textbf{(D)} \: 11{,}000\qquad\textbf{(E)} \: 11{,}110$

Solution 1

We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$, we find that the sum is equal to \[10\cdot(1+10+100+1000)=\boxed{\textbf{(E)} \: 11{,}110}.\] Note that it is equally valid to manually add all four numbers together to get the answer.

~kingofpineapplz

Solution 2

We have \[1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.\] ~Steven Chen (www.professorchenedu.com)

Solution 3

We see that the units digit must be $0$, since $4+3+2+1$ is $0$. But every digit from there, will be a $1$ since we have that each time afterwards, we must carry the $1$ from the previous sum. The answer choice that satisfies these conditions is $\boxed{\textbf{(E)} \: 11{,}110}$.

~stjwyl

Solution 4 (Brute Force)

We can simply add the numbers. $1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{\textbf{(E)}}$.

Note: Although this would not take terribly long, it is not recommended to do this in a real contest.

~ cxsmi

Video Solution by Interstigation

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

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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