Difference between revisions of "2003 AMC 12B Problems/Problem 8"

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==Solution==
 
==Solution==
Let <math>a</math> and <math>b</math> be the digits of <math>\clubsuit(x)</math>,
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Let <math>y=\clubsuit (x)</math>. Since <math>x \leq 99</math>, we have <math>y \leq 18</math>. Thus if <math>\clubsuit (y)=3</math>, then <math>y=3</math> or <math>y=12</math>. The 3 values of <math>x</math> for which <math>\clubsuit (x)=3</math> are 12, 21, and 30, and the 7 values of <math>x</math> for which <math>\clubsuit (x)=12</math> are 39, 48, 57, 66, 75, 84, and 93. There are <math>\boxed{E=10}</math> values in all.
 
 
<cmath>\clubsuit(\clubsuit(x)) = a + b = 3</cmath>
 
 
 
Clearly <math>\clubsuit(x)</math> can only be <math>3, 12, 21,</math> or <math>30</math> and only <math>3</math> and <math>12</math> are possible to have two digits sum to.
 
 
 
If <math>\clubsuit(x)</math> sums to <math>3</math>, there are 3 different solutions : <math>12, 21, \text{or } 30</math>  
 
 
 
If <math>\clubsuit(x)</math> sums to <math>12</math>, there are 7 different solutions: <math>39, 48, 57, 66,75, 84, \text{or } 93</math>
 
 
 
The total number of solutions is <math> 3 + 7 =10 \Rightarrow \text (E)</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 15:38, 30 March 2020

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 $y=\clubsuit (x)$. Since $x \leq 99$, we have $y \leq 18$. Thus if $\clubsuit (y)=3$, then $y=3$ or $y=12$. The 3 values of $x$ for which $\clubsuit (x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\clubsuit (x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\boxed{E=10}$ values in all.

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions
2003 AMC 10B (ProblemsAnswer KeyResources)
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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