Bunuel wrote:

Which of the following CANNOT result in an integer?

A. The product of two integers divided by the reciprocal of a different integer

B. An even integer divided by 7

C. The quotient of two distinct prime numbers

D. A multiple of 11 divided by 3

E. The sum of two odd integers divided by 2

\(?\,\,:\,\,\,{\rm{cannot}}\,\,{\rm{be}}\,\,{\rm{integer}}\)

\(\left( {\rm{A}} \right)\,\,\,\left( {0 \cdot 0} \right) \div {1^{ - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{A}} \right)\,\,\,{\rm{refuted}}\)

\(\left( {\rm{B}} \right)\,\,\,0 \div 7\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{B}} \right)\,\,\,{\rm{refuted}}\)

\(\left( {\rm{D}} \right)\,\,0 \div 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)\,\,\,{\rm{refuted}}\)

\(\left( {\rm{E}} \right)\,\,\left( {1 + 1} \right) \div 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{E}} \right)\,\,\,{\rm{refuted}}\)

Conclusion: (C) is the corrrect answer by exclusion.

POST-MORTEM:

\(\left( {\rm{C}} \right)\,\,\,{{{p_1}} \over {{p_2}}} \ne {\mathop{\rm int}} \,\,\,\,:\,\,\,\,\,{\rm{if}}\,\,\,\,{{{p_1}} \over {{p_2}}} = {\mathop{\rm int}} \,\,\,\,\left\{ \matrix{

{p_1}\,\,,{p_2}\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\,\left( * \right) \hfill \cr

{p_1} \ne {p_2}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{\mathop{\rm int}} \,\, \ge 2 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{p_1} = {\mathop{\rm int}} \,\, \cdot {p_2}\,\,\,\,\,\,\,\,\)

\(\,\mathop \Rightarrow \limits_{{p_2}\,\, \ge \,\,2}^{{\mathop{\rm int}} \,\, \ge \,\,2} \,\,\,\,\,\,\,\,{p_1}\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\)

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)

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