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

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

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 1: / Line 1: @@
-{{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]]}}
+{{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]]}}
 ==Problem==
-What is the value of <math>1234+2341+3412+4123?</math>
+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>
 == 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.
+~kingofpineapplz
-Note: it is equally valid to manually add all 4 numbers together to get the answer.
+== Solution 2 ==
+We have <cmath>1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.</cmath>
+~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 <math>1</math> from the previous sum. The answer choice that satisfies these conditions is <math>\boxed{\textbf{(E)} \: 11{,}110}</math>.
-~kingofpineapplz
+~stjwyl
-== Solution 2 ==
+== Solution 4 (Brute Force)==
-We have
+We can simply add the numbers. <math>1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{\textbf{(E)}}</math>.
-<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>.
+Note: Although this would not take terribly long, it is not recommended to do this in a real contest.
-~Steven Chen (www.professorchenedu.com)
+~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
-== 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>.
-~~stjwyl
 ==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==

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