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

Revision as of 19:01, 20 December 2012

Let $a$ and $b$ be the digits of $x$ ,

$\clubsuit(\clubsuit(x)) = a + b = 3$

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 : $12, 21, or 30$

If $\clubsuit(x)$ sums to $12$ , there are 7 different solutions: $39, 48, 57, 66,75, 84, or 93$

The total number of solutions is $3 + 7 =10 \Rightarrow \text (E)$

@@ Line 3: / Line 3: @@
 <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>30</math> are possible to have two digits sum to.
+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> is \Rightarrow \text (E)$
+If <math>\clubsuit(x)</math> sums to <math>3</math>, there are 3 different solutions : <math>12, 21, 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, or 93</math>
+The total number of solutions is <math> 3 + 7 =10 \Rightarrow \text (E)</math>