# 2021 Fall AMC 12B Problems/Problem 1

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.

## Video Solution

~Education, the Study of Everything

~savannahsolver