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2003 AMC 10B Problem 13

2003 AMC 10B Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 2 of 5 · about 3 min · position P11-15

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 $\clubsuit(\clubsuit(x))=3$ ?

A)
$3$
B)
$4$
C)
$6$
D)
$9$
E)
$10$

Answer

Key insight

The digit sum of a two-digit number is at most 18, so it must be 3 or 12; count numbers with those digit sums: 3 + 7.

Solution

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 $\boxed{\textbf{(E)}\ 10}$ .

Why this works

Iterated digit sums shrink fast, so working from the outside in leaves only a couple of possible intermediate values. Bounding the inner quantity first ( $s \le 18$ ) converts a vague two-layer condition into two concrete one-layer counts, each of which is a short list indexed by the tens digit.

The trap

Forgetting the digit sum 12 case (answering 3), or including 03 as a two-digit number with digit sum 3.

Common mistakes

Forgetting the digit sum $12$ case (answering $3$ ), or including $03$ as a two-digit number with digit sum $3$ .
Listing the digit-sum- $12$ numbers with tens digit starting at $2$ (giving $29$ , which sums to $11$ ) or missing $66$ .

Techniques

Organized listing / direct enumeration · Split into exhaustive cases and handle each