Difference between revisions of "2021 Fall AMC 12B Problems/Problem 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>
 
<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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~stjwyl
 
~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]
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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https://youtu.be/I9lSM0hO39M
 
https://youtu.be/I9lSM0hO39M
  
~Education, the Study of Education
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~Education, the Study of Everything
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==
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~savannahsolver
 
~savannahsolver
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/lC7naDZ1Eu4
 +
 +
For AMC 12: https://youtu.be/yaE5aAmeesc
 +
 +
~IceMatrix
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==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}}

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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