Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
\(1 \leqslant n \leqslant 179\,\,\,\,\,\left( {n\,\,\,\operatorname{int} } \right)\,\,\,\,\left( * \right)\)
\(n = 13M + 2\,\,\,\,,\,\,\,M\,\,\operatorname{int} \,\,\,\,\left( {\text{I}} \right)\)
\(n = 8J + 5\,\,\,\,,\,\,\,J\,\,\operatorname{int} \,\,\,\,\left( {{\text{II}}} \right)\)
\(?\,\,\,:\,\,\,n\,\,\,{\text{in}}\,\,\,\left( * \right) \cap \left( {\text{I}} \right) \cap \left( {{\text{II}}} \right)\)
\(\left( * \right)\,\, \cap \,\,\left( {\text{I}} \right)\,\,\,:\,\,\,\,\,\,1\,\, \leqslant \,\,13M + 2\,\, \leqslant \,\,179\,\,\,\,\,\mathop \Leftrightarrow \limits^{ - \,2} \,\,\,\,\, - 1 \leqslant 13M \leqslant 177\,\,\left( { = 169 + 8} \right)\)
\(- 1 \leqslant 13M \leqslant 177\,\,\left( { = 169 + 8} \right)\,\,\,\,\,\mathop \Leftrightarrow \limits^{M\,\,\operatorname{int} } \,\,\,0 \leqslant 13M \leqslant 169\,\,\,\,\,\mathop \Leftrightarrow \limits^{:\,\,13} \,\,\,\,0 \leqslant M \leqslant 13\)
\(\left( * \right)\,\, \cap \,\,\left( {{\text{II}}} \right)\,\,\,:\,\,\,\,\,\,\left\{ \begin{gathered}
\,n - 5\,\,{\text{divisible}}\,\,{\text{by}}\,\,8\,\,\,\,\, \Rightarrow \,\,\,\,\,n - 5\,\,\,\,\,{\text{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\,n\,\,\,{\text{odd}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {\text{I}} \right)} \,\,\,\,\,M\,\,{\text{odd}} \hfill \\
\,n - 5\mathop = \limits^{\left( {\text{I}} \right)} 13M - 3\,\,{\text{divisible}}\,\,{\text{by}}\,\,8\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{13M - 3}}{2}\,\,\,\,{\text{divisible}}\,\,{\text{by}}\,\,4\,\,\,\left( {***} \right)\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\)
\(\left. \begin{gathered}
M = 1\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 - 3}}{2} = 5\,\,\,{\text{odd}}\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 3\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 3 - 3}}{2} = 18\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 5\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 5 - 3}}{2} = 31\,\,{\text{odd}}\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}}\,\,\,\,\,\, \hfill \\
\boxed{M = 7}\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 7 - 3}}{2} = 44\,\,\,\,\left( {***} \right)\,\,\,{\text{YES}} \hfill \\
M = 9\,\,\,\,\, \Rightarrow \,\,\,\,{\text{odd}}\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 11\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 11 - 3}}{2} = 70\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 13\,\,\,\,\, \Rightarrow \,\,\,\,{\text{odd}}\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
\end{gathered} \right\}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 1\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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