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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>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.
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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}</math> = <math>\boxed{10}</math> values in all.
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~Andrew_Lu/Mismatchedcubing
  
 
==See Also==
 
==See Also==

Latest revision as of 11:40, 1 August 2024

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 $\boxed{E}$ = $\boxed{10}$ values in all.

~Andrew_Lu/Mismatchedcubing

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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