Printable

2016 AMC 10A Problem 6

2016 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 2 of 5 · about 2 min · position P1-10

Ximena lists the whole numbers $1$ through $30$ once. Emilio copies Ximena's numbers, replacing each occurrence of the digit $2$ by the digit $1$ . Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?

A)
$13$
B)
$26$
C)
$102$
D)
$103$
E)
$110$

Answer

Key insight

Each replaced units digit 2 lowers the sum by 1 and each replaced tens digit 2 lowers it by 10; count the digit positions, not the numbers.

Solution

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 $\boxed{\textbf{(D)}\ 103}$ .

Why this works

Place value makes the effect of a digit change independent of the rest of the number, so the total difference is a sum over digit positions. Count occurrences of the digit in each place separately and weight by the place value; do not try to list the modified numbers.

The trap

Counting only the units-digit 2s (giving 3) or treating every number containing a 2 as losing exactly 1 or 10, and missing that 22 loses 11.

Common mistakes

Counting only the units-digit 2s (giving 3) or treating every number containing a 2 as losing exactly 1 or 10, and missing that 22 loses 11.
Counting the tens-place $2$ s as $20$ through $30$ ( $11$ numbers) instead of $20$ through $29$ , which gives $113$ .

Techniques

Set up the equation/formula and compute; no special trick needed