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

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==Solution==
 
==Solution==
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Let <math>a</math> and <math>b</math> be the digits of <math>x</math>,
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Let <math>a</math> and <math>b</math> be the digits of <math>\clubsuit(x)</math>,
  
 
<cmath>\clubsuit(\clubsuit(x)) = a + b = 3</cmath>
 
<cmath>\clubsuit(\clubsuit(x)) = a + b = 3</cmath>

Revision as of 17:31, 30 November 2019

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 $\clubsuit(x)$,

\[\clubsuit(\clubsuit(x)) = a + b = 3\]

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 $3 + 7 =10 \Rightarrow \text (E)$

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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