2003 AMC 12B Problems/Problem 8

The following problem is from both the 2003 AMC 12B #8 and 2003 AMC 10B #13, so both problems redirect to this page.

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

Let $a$ and $b$ be the digits of $x$ ,

$\clubsuit(\clubsuit(x)) = a + b = 3$

Clearly $\clubsuit(x)$ can only be $3, 12, 21,$ or $30$ and only $3$ and $12$ are possible to have two digits sum to.

If $\clubsuit(x)$ sums to $3$ , there are 3 different solutions : $12, 21, \text{or } 30$

If $\clubsuit(x)$ sums to $12$ , there are 7 different solutions: $39, 48, 57, 66,75, 84, \text{or } 93$

The total number of solutions is $3 + 7 =10 \Rightarrow \text (E)$

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 10 Problems and Solutions

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2003 AMC 12B Problems/Problem 8

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