2016 AMC 10A Problem 6

P1-10~2 minPrint

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

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?

Show solution without answering

Key insight

A 2 turned into 1 costs 1 in the ones place and 10 in the tens place. Count the 2s in each place separately.

Solution

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

If a $2$ in the ones place becomes a $1$ , the number drops by $1$ . For example, $12$ becomes $11$ .
If a $2$ in the tens place becomes a $1$ , the number drops by $10$ . For example, $25$ becomes $15$ .
Ones-place $2$ s from $1$ to $30$ : the numbers $2$ , $12$ and $22$ . That is $3$ digits, costing $3 \cdot 1 = 3$ .
Tens-place $2$ s: the numbers $20, 21, \ldots, 29$ . That is $10$ digits, costing $10 \cdot 10 = 100$ .
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 $\boxed{\textbf{(D)}\ 103}$ .

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Number Theory, Chapter 9: Base Number Arithmetic
Introduction to Number Theory, Chapter 6: Special Numbers (§6.5)