Positive integers A, B, C are such that C is a multiple of B & B is a

\(A,B,C\,\,\, \geqslant 1\,\,\,{\text{ints}}\)

\(\frac{C}{B} = \operatorname{int} \,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\frac{B}{A} = \operatorname{int}\)

\(\frac{A}{5}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int}\)

\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{\\
{{BC} \over 5} = {\mathop{\rm int}} \hfill \cr \\
{{AB} \over 5} = {\mathop{\rm int}} \hfill \cr} \right.\,\,\,\,\,\,\,\)

\(\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {1,5,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {5,5,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)


The correct answer is therefore (E).


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

Regards,
Fabio.