2016 AMC 10A Problem 6

Instead of rewriting all thirty numbers, ask how much each changed digit costs.

1. If a $2$ in the ones place becomes a $1$, the number drops by $1$. For example, $12$ becomes $11$.
2. If a $2$ in the tens place becomes a $1$, the number drops by $10$. For example, $25$ becomes $15$.
3. Ones-place $2$s from $1$ to $30$: the numbers $2$, $12$ and $22$. That is $3$ digits, costing $3 \cdot 1 = 3$.
4. Tens-place $2$s: the numbers $20, 21, \ldots, 29$. That is $10$ digits, costing $10 \cdot 10 = 100$.
5. The number $22$ shows up in both lists, and that is correct: it becomes $11$, so it really drops by $11 = 1 + 10$.

Ximena's sum is larger by $3 + 100 = 103$.

The answer is $\boxed{\textbf{(D)}\ 103}$.

Place value means each digit's contribution to a number is separate from the others: a digit in the tens place is worth ten times its face value no matter what the ones digit is. So the total change is a sum over digit positions. Count how many times the digit appears in each place and multiply by the place value.

Work digit by digit. Changing a $2$ to a $1$ in the units place lowers a number by $1$; changing a $2$ to a $1$ in the tens place lowers it by $10$.

Units-place $2$s among $1$ through $30$: the numbers $2$, $12$, $22$, three of them. Total loss: $3 \cdot 1 = 3$.

Tens-place $2$s: the numbers $20$ through $29$, ten of them. Total loss: $10 \cdot 10 = 100$.

(The number $22$ appears in both counts, and correctly so: it loses $11$.)

Ximena's sum exceeds Emilio's by $3 + 100 = 103$. The answer is $\boxed{\textbf{(D)}\ 103}$.