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

Revision as of 09:25, 4 July 2013

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 $3 + 7 =10 \Rightarrow \text (E)$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 18:01, 20 December 2012 (view source) Mathematician153 (talk \| contribs) ← Older edit		Revision as of 09:25, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	The total number of solutions is <math> 3 + 7 =10 \Rightarrow \text (E)</math>		The total number of solutions is <math> 3 + 7 =10 \Rightarrow \text (E)</math>
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