2003 AMC 10B Problems/Problem 13

Problem

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

$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10$

Solution

We can divide $\clubsuit(x)$ into two cases so that $\clubsuit(\clubsuit(x))=3.$ The first is where $\clubsuit(x)$ is a one-digit number, and the second is where it is a two=digit number.

For $\clubsuit(x)$ to be a one-digit number, $x$ 's digits must add up to be $3.$ This can be done in three ways $\longrightarrow 30, 21,$ and $12.$

For $\clubsuit(x)$ to be a two-digit number, $x$ 's digits must add up to be $12,$ since the sum cannot exceed $9+9=18.$ This can be done in seven ways $\longrightarrow 93, 84, 75, 66, 57, 48,$ and $39.$

Add the number of ways together $\longrightarrow 3+7=\boxed{\mathrm{(E) \ } 10}$

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2003 AMC 10B Problems/Problem 13

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