BabySmurf wrote:

Is positive integer x divisible by 24?

(1) √x is divisible by 4.

(2) x^2 is not divisible by 9.

\(x \geqslant 1\,\,\operatorname{int}\)

\(\frac{x}{{{2^3} \cdot 3}}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int}\)

\(\left( 1 \right)\,\,\,\,\frac{{\sqrt x }}{{{2^2}}}\,\,\, = \,\,\,\operatorname{int} \,\,\,\,\,\,\left\{ \begin{gathered}

\,Take\,\,x = {2^4}\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\

\,Take\,\,x = {\left( {{2^3} \cdot 3} \right)^2}\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\

\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,{\left( {\frac{x}{3}} \right)^2} = \frac{{{x^2}}}{{{3^2}}}\,\,\, \ne \,\,\,\operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{3}\,\,\, \ne \,\,\,\operatorname{int} \,\,\,\,\,\,\,\mathop \Rightarrow \limits^{FOCUS\,!} \,\,\,\,\,\frac{x}{3}\,\,\, \ne \,\,\,{2^3} \cdot \operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{{{2^3} \cdot 3}}\,\,\, \ne \,\,\,\operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)

The above follows the notations and rationale taught in the GMATH method.

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Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)

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