AAPL wrote:GMAT Prep
There are 27 different three-digit integers that can be formed using only the digits 1, 2, and 3. If all 27 of the integers were listed, what would their sum be?
A. 2,704
B. 2,990
C. 5,404
D. 5,444
E. 5,994
\[? = 111 + 112 + 113 + 121 + 122 + 123 + \ldots + 331 + 332 + 333\]
\[1 \,\, \, \underline {} \, \, \, \underline {} :\,111 + 112 + 113 + 121 + 122 + 123 + 131 + 132 + 133\]
\[\left\{ \begin{gathered}
\,1 \to {\text{9}}\,\,{\text{times}}\,\,{\text{in}}\,\,{\text{hundreds}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{tens}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{units}}\,\,{\text{digit}} \hfill \\
\end{gathered} \right.\]
\[{\text{Same}}\,\,{\text{occurs}}\,\,{\text{with}}\,\,2 \,\,\, \underline {} \,\,\, \underline {} \, \,\,\, {\text{and}} \,\,\, \,\,3 \,\,\, \underline {} \,\,\, \underline {} \,\,,\,\,{\text{hence:}}\]
\[{\text{?}}\,\,\,{\text{:}}\,\,\,\left\{ \begin{gathered}
\,1,2,3 \to {\text{9}}\,\,{\text{times}}\,\,{\text{in}}\,\,{\text{hundreds}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3 + 3 + 3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{tens}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3 + 3 + 3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{units}}\,\,{\text{digit}} \hfill \\
\end{gathered} \right.\]
\[? = 9\left( {1 + 2 + 3} \right) \cdot 100 + 9\left( {1 + 2 + 3} \right) \cdot 10 + 9\left( {1 + 2 + 3} \right)\]
\[? = 9 \cdot 6 \cdot \left( {100+10+1} \right) = 5994\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
