Stiv wrote:
Is the three-digit number n less than 550?
(1) The product of the digits in n is 30.
(2) The sum of the digits in n is 10.
\(100 \leqslant \,\,n = \left\langle {ABC} \right\rangle \,\, \leqslant 999\)
\(\left\langle {ABC} \right\rangle \,\,\mathop < \limits^? \,\,550\)
\(\left( 1 \right)\,\,\,A \cdot B \cdot C = 30\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {1,5,6} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {5,6,1} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,\,A + B + C = 10\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {1,2,7} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {5,5,0} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)
\(\left( {1 + 2} \right)\,\,\,\left\langle {ABC} \right\rangle \,\,\, \geqslant \,\,\,550\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}\\
\,A = 5\,\,,\,\,B \geqslant 5\,\,\,\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,B = 5,\,\,C = 0\,\,\, \Rightarrow \,\,\,\left( 1 \right)\,\,{\text{contradicted}} \hfill \\\\
\,\,\,{\text{or}} \hfill \\\\
\,A \geqslant 6\,\,\,\mathop \Rightarrow \limits_{\left( 1 \right)}^{\left( * \right)} \,\,\,A = 6\,\,\,\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,B \cdot C = 5\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,A + B + C = 12\,\,\, \Rightarrow \,\,\,\left( 2 \right)\,\,{\text{contradicted}} \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)
\(\left( * \right)\,\,\,\left( 1 \right)\,\,\, \Rightarrow \,\,\,A,B,C\,\,{\text{are}}\,\,\left( {{\text{positive}}} \right)\,\,{\text{divisors}}\,\,{\text{of}}\,\,30\,\,\, \Rightarrow \,\,\,A \notin \left\{ {7,8,9} \right\}\)
\(\left( {**} \right)\,\,\,B \cdot C = 5\,\,\, \Rightarrow \,\,\,B,C\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{pair}}\,\,{\text{of}}\,\,\left( {{\text{positive}}} \right)\,\,{\text{divisors}}\,\,{\text{of}}\,\,5\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}\\
\,\left( {B,C} \right) = \left( {1,5} \right) \hfill \\\\
\,\,{\text{or}} \hfill \\\\
\,\left( {B,C} \right) = \left( {5,1} \right) \hfill \\ \\
\end{gathered} \right.\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)