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

(Solution 1)
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~kingofpineapplz
 
~kingofpineapplz
 +
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== Solution 2 ==
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We have
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<cmath>
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\begin{align*}
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1234 + 2341 + 3412 + 4123
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& = 1111 \left( 1 + 2 + 3 + 4 \right) \\
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& = 11110 .
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\end{align*}
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</cmath>
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Therefore, the answer is <math>\boxed{\textbf{(E) }11,110}</math>.
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~Steven Chen (www.professorchenedu.com)
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==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/p9_RH4s-kBA
 
https://youtu.be/p9_RH4s-kBA

Revision as of 21:13, 25 November 2021

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: it is equally valid to manually add all 4 numbers together to get the answer.


~kingofpineapplz

Solution 2

We have \begin{align*} 1234 + 2341 + 3412 + 4123 & = 1111 \left( 1 + 2 + 3 + 4 \right) \\ & = 11110 . \end{align*}

Therefore, the answer is $\boxed{\textbf{(E) }11,110}$.

~Steven Chen (www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA

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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