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Difference between revisions of "2003 AMC 10B Problems/Problem 13"

(Created page with "==Problem== 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)=...")
 
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==Solution==
 
==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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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>\longrightarrow 30, 21,</math> and <math>12.</math>
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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>\longrightarrow 93, 84, 75, 66, 57, 48,</math> and <math>39.</math>
+
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>\longrightarrow 3+7=\boxed{\mathrm{(E) \ } 10}</math>
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Add the number of ways together <math>\Rightarrow 3+7=\boxed{\textbf{(E)}\ 10}</math>
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2003|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2003|ab=B|num-b=12|num-a=14}}

Revision as of 19:18, 26 November 2011

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 $\Rightarrow 30, 21,$ 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 $\Rightarrow 93, 84, 75, 66, 57, 48,$ and $39.$

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

See Also

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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