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| − | ==Problem==
| + | #REDIRECT [[2003 AMC 12B Problems/Problem 8]] |
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| − | Let <math>\clubsuit(x)</math> denote the sum of the digits of the positive integer <math>x</math>. For example, <math>\clubsuit(8)=8</math> and <math>\clubsuit(123)=1+2+3=6</math>. For how many two-digit values of <math>x</math> is <math>\clubsuit(\clubsuit(x))=3</math>?
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| − | <math>\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10 </math>
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| − | ==Solution==
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| − | We can divide <math>\clubsuit(x)</math> into two cases so that <math>\clubsuit(\clubsuit(x))=3.</math> The first is where <math>\clubsuit(x)</math> is a one-digit number, and the second is where it is a two-digit number.
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| − | For <math>\clubsuit(x)</math> to be a one-digit number, <math>x</math>'s digits must add up to be <math>3.</math> This can be done in three ways <math>\Rightarrow 30, 21,</math> and <math>12.</math>
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| − | For <math>\clubsuit(x)</math> to be a two-digit number, <math>x</math>'s digits must add up to be <math>12,</math> since the sum cannot exceed <math>9+9=18.</math> This can be done in seven ways <math>\Rightarrow 93, 84, 75, 66, 57, 48,</math> and <math>39.</math>
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| − | Add the number of ways together <math>\Rightarrow 3+7=\boxed{\textbf{(E)}\ 10}</math>
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| − | ==See Also==
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| − | {{AMC10 box|year=2003|ab=B|num-b=12|num-a=14}}
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| − | {{MAA Notice}}
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