2003 AMC 10B Problem 13

P11-15~3 minPrint

Bases & Digits Basic Counting Organized listing / direct enumeration Split into exhaustive cases and handle each

Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x$ . For example, $\clubsuit(8)=8$ and $\clubsuit(123)=1+2+3=6$ . For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x))=3$ ?

Show solution without answering

Key insight

A two-digit number's digits add to at most 18, and only 3 and 12 in that range have digit sum 3. Count both cases.

Solution

Work from the outside in. Let $s$ be the digit sum of $x$ . The question needs the digit sum of $s$ to be $3$ .
How big can $s$ get? The biggest two-digit number is $99$ , with digit sum $9 + 9 = 18$ . So $s$ is between $1$ and $18$ .
Which numbers from $1$ to $18$ have digit sum $3$ ? Among $1$ to $9$ , only $3$ . Among $10$ to $18$ , the digit sum is $1$ plus the last digit, which is $3$ only for $12$ . So $s = 3$ or $s = 12$ .
Two-digit numbers with digit sum $3$ : $12, 21, 30$ . That is $3$ numbers.
Two-digit numbers with digit sum $12$ : the tens digit must be at least $3$ (otherwise the units digit would need to be more than $9$ ). So we get $39, 48, 57, 66, 75, 84, 93$ . That is $7$ numbers.
Total: $3 + 7 = 10$ .

The answer is $\boxed{\textbf{(E)}\ 10}$ .

Why this works

Taking a digit sum makes a number much smaller, so the inner digit sum can only be a few values. Finding the largest it can be (here $18$ ) turns a confusing "digit sum of a digit sum" condition into two short lists. Listing by tens digit makes sure no number is missed.

Concise solution

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

For a two-digit $x$ , the digit sum $s = \clubsuit(x)$ lies between $1$ and $9 + 9 = 18$ . We need $\clubsuit(s) = 3$ with $1 \le s \le 18$ . Single-digit $s$ works only for $s = 3$ ; two-digit $s$ from $10$ to $18$ has digit sum $1 + (\text{units digit})$ , which equals $3$ only for $s = 12$ .

Now count two-digit $x$ with each digit sum.

Digit sum $3$ : tens digit $1, 2, 3$ with units $2, 1, 0$ : $12, 21, 30$ . That is $3$ numbers.
Digit sum $12$ : tens digit $t$ from $3$ to $9$ with units $12 - t$ from $9$ down to $3$ : $39, 48, 57, 66, 75, 84, 93$ . That is $7$ numbers.

Total: $3 + 7 = 10$ .

The answer is $\boxed{\textbf{(E)}\ 10}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)