Problem Solving

Veritas Prep

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

OA B

BTGmoderatorLU wrote:Veritas Prep

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

$$1 \le n \le 179\,\,\,\,\,\left( {n\,\,\,{\mathop{\rm int}} } \right)\,\,\,\,\left( * \right)$$
$$n = 13M + 2\,\,\,\,,\,\,\,M\,\,{\mathop{\rm int}} \,\,\,\,\left( {\rm{I}} \right)$$
$$n = 8J + 5\,\,\,\,,\,\,\,J\,\,{\mathop{\rm int}} \,\,\,\,\left( {{\rm{II}}} \right)$$
\[?\,\,\,:\,\,\,n\,\,\,{\text{in}}\,\,\,\left( * \right) \cap \left( {\text{I}} \right) \cap \left( {{\text{II}}} \right)\]

$$\left( * \right)\,\, \cap \,\,\left( {\rm{I}} \right)\,\,\,:\,\,\,\,\,\,1\,\, \le \,\,13M + 2\,\, \le \,\,179\,\,\,\,\,\mathop \Leftrightarrow \limits^{ - \,2} \,\,\,\,\, - 1 \le 13M \le 177\,\,\left( { = 169 + 8} \right)$$
$$ - 1 \le 13M \le 177\,\,\left( { = 169 + 8} \right)\,\,\,\,\,\mathop \Leftrightarrow \limits^{M\,\,{\mathop{\rm int}} } \,\,\,0 \le 13M \le 169\,\,\,\,\,\mathop \Leftrightarrow \limits^{:\,\,13} \,\,\,\,0 \le M \le 13$$

$$\left( * \right)\,\, \cap \,\,\left( {{\rm{II}}} \right)\,\,\,:\,\,\,\,\,\,\left\{ \matrix{
\,n - 5\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,8\,\,\,\,\, \Rightarrow \,\,\,\,\,n - 5\,\,\,\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\,n\,\,\,{\rm{odd}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {\rm{I}} \right)} \,\,\,\,\,M\,\,{\rm{odd}} \hfill \cr
\,n - 5\mathop = \limits^{\left( {\rm{I}} \right)} 13M - 3\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,8\,\,\,\,\, \Rightarrow \,\,\,\,\,{{13M - 3} \over 2}\,\,\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,4\,\,\,\left( {***} \right)\,\,\,\,\,\, \hfill \cr} \right.$$
\[\left. \begin{gathered}
M = 1\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 - 3}}{2} = 5\,\,\,{\text{odd}}\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 3\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 3 - 3}}{2} = 18\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 5\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 5 - 3}}{2} = 31\,\,{\text{odd}}\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}}\,\,\,\,\,\, \hfill \\
\boxed{M = 7}\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 7 - 3}}{2} = 44\,\,\,\,\left( {***} \right)\,\,\,{\text{YES}} \hfill \\
M = 9\,\,\,\,\, \Rightarrow \,\,\,\,{\text{odd}}\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 11\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 11 - 3}}{2} = 70\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
M = 13\,\,\,\,\, \Rightarrow \,\,\,\,{\text{odd}}\,\,\,\,\left( {***} \right)\,\,\,{\text{NO}} \hfill \\
\end{gathered} \right\}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 1\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.