2021 AMC Fall 10B Problem 1

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

The four summands are cyclic shifts of the same digit string, so the sum is symmetric across place values. Whenever a sum has this rotational structure, factor it as (digit sum) times a repunit like $1111$ instead of adding column by column; it is faster and avoids carrying errors.

Plain column addition: units $4+1+2+3 = 10$, 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 $1+2+3+4+1 = 11$. Result $11110$.

• Adding the four numbers column by column and slipping on a carry, landing on 10,110 or 11,000.
• Multiplying $10 \cdot 1111$ and writing $10{,}110$ instead of $11{,}110$.