EgmatQuantExpert wrote:
The number A can be expressed as p x q, where p and q are positive integers. Is A divisible by 16?
1. p = 8 x k, where k is an odd number.
2. \(q^2 – 8q + 15 = 0\)
\(p,q\,\, \geqslant 1\,\,\,{\text{ints}}\)
\(\frac{{pq}}{{{2^4}}}\,\,\mathop = \limits^? \,\,\operatorname{int}\)
\(\left( 1 \right)\,\,\,p = 8k\,\,,\,\,\,k\,\,{\text{odd}}\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {p,q} \right) = \left( {8,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {p,q} \right) = \left( {8,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,\,\left( {q - 3} \right)\left( {q - 5} \right) = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,q = 3\,\,\,{\text{or}}\,\,\,q = 5\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {p,q} \right) = \left( {8,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {p,q} \right) = \left( {16,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\)
\(\left( {1 + 2} \right)\,\,\,pq = {2^3} \cdot M\,,\,\,M\,\,{\text{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)
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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