Max@Math Revolution wrote:[GMAT math practice question]
If x, y, z are positive integers, is xyz ≥ 64?
1) xy ≥ yz ≥ zx ≥ 16
2) x + y + z = 64
Excellent problem, Max. Congrats!
$$x,y,z\,\, \ge 1\,\,{\rm{ints}}$$
$$xyz\,\,\mathop \ge \limits^? \,\,64$$
$$\left( 1 \right)\,\,xy \ge yz \ge xz \ge 16$$
$$\left\{ \matrix{
\,\left( {yz} \right)x \ge 16x \hfill \cr
\,\left( {xz} \right)y \ge 16y \hfill \cr
\,\left( {xy} \right)z \ge 16z \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits_{\left( {{\rm{all}}\,\,{\rm{positive}}} \right)}^{multiply\,\,!} \,\,\,\,{\left( {xyz} \right)^3} \ge {16^3}\left( {xyz} \right)\,\,\,\,\,\,\,\left( * \right)$$
$$\left( * \right)\,\,\,\,\mathop \Rightarrow \limits_{\left( {xyz\,\,{\rm{positive}}} \right)}^{:\,\,xyz} \,\,\,\,\,{\left( {xyz} \right)^2} \ge {16^3}\,\,\,\,\mathop \Rightarrow \limits^{xyz\,\, > \,\,0} \,\,\,\,xyz \ge \sqrt {{2^{12}}} = 64\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
$$\left( 2 \right)\,\,\,x + y + z = 64\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( {1,1,62} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( {2,2,60} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \, \hfill \cr} \right.$$
The correct answer is therefore (A).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
