2021 AMC Fall 10B Problem 1

Before adding, stack the numbers and look at the columns. There may be a pattern that saves work.

$$\begin{array}{r} 1234 \\ 2341 \\ 3412 \\ 4123 \end{array}$$

1. Thousands column: $1, 2, 3, 4$.
2. Hundreds column: $2, 3, 4, 1$.
3. Tens column: $3, 4, 1, 2$.
4. Ones column: $4, 1, 2, 3$.

Every column holds the same four digits $1, 2, 3, 4$, just in a different order. So every column adds up to $1 + 2 + 3 + 4 = 10$.

5. That means the total is $10$ thousands, plus $10$ hundreds, plus $10$ tens, plus $10$ ones:
$$10 \times 1000 + 10 \times 100 + 10 \times 10 + 10 \times 1 = 10 \times 1111 = 11{,}110.$$

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

Each number is the one before it with its first digit moved to the end. Because of that, every digit visits every place value exactly once. When you see numbers built like this, add the digits once and multiply by $1111$ (or $111$, and so on) instead of carrying column by column.

Ordinary column addition also works if you are careful with carries.

• Ones: $4 + 1 + 2 + 3 = 10$. Write $0$, carry $1$.
• Tens: $3 + 4 + 1 + 2 + 1 = 11$. Write $1$, carry $1$.
• Hundreds: $2 + 3 + 4 + 1 + 1 = 11$. Write $1$, carry $1$.
• Thousands: $1 + 2 + 3 + 4 + 1 = 11$. Write $11$.

The result is $11{,}110$.

Look at the four numbers as a block of digits. In the thousands place the digits are $1, 2, 3, 4$; in the hundreds place they are $2, 3, 4, 1$; in the tens place $3, 4, 1, 2$; in the units place $4, 1, 2, 3$. Every place value receives each of the digits $1, 2, 3, 4$ exactly once, so every column sums to $1 + 2 + 3 + 4 = 10$.

Therefore the total is
$$10 \cdot 1000 + 10 \cdot 100 + 10 \cdot 10 + 10 \cdot 1 = 10 \cdot 1111 = 11{,}110.$$

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