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

Revision as of 18:22, 29 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)

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. The answer choice that satisfies these conditions is $\boxed{\textbf{(E) }11,110}$ .~~stjwyl

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA

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.

@@ Line 28: / Line 28: @@
 ~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 1. The answer choice that satisfies these conditions is <math>\boxed{\textbf{(E) }11,110}</math>.~~stjwyl
 ==Video Solution by Interstigation==
 https://youtu.be/p9_RH4s-kBA

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