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

Revision as of 05:33, 3 January 2022

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

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 7: / Line 7: @@
 == 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>
+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.
-Note: it is equally valid to manually add all 4 numbers together to get the answer.
 ~kingofpineapplz
 == Solution 2 ==
-We have
+We have <cmath>1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.</cmath>
-<cmath>
-\begin{align*}
-+ 2341 + 3412 + 4123
-& = 1111 \left( 1 + 2 + 3 + 4 \right) \\
-& = 11110 .
-\end{align*}
-</cmath>
-Therefore, the answer is <math>\boxed{\textbf{(E) }11,110}</math>.
 ~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 from the previous sum. The answer choice that satisfies these conditions is <math>\boxed{\textbf{(E) }11,110}</math>.
+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 from the previous sum. 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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