fiza gupta wrote:If x and y are positive integers, is the total number of positive divisors of x^3 a multiple of the total number of positive divisors of y^2?
(1) x=4
(2) y=6
\[x,y\,\, \geqslant 1\,\,\,{\text{ints}}\]
\[\frac{{\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\,}}{{\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{y^2}}}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} \]
\[\left( 1 \right)\,\,\,x = {2^2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\left( { = {2^{\,6}}} \right)\,\,\, = \,\,\,7\]
\[\left\{ \begin{gathered}
\,{\text{Take}}\,\,y = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\left( {\frac{7}{1} = \operatorname{int} } \right)\,\, \hfill \\
\,{\text{Take}}\,\,y = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\left( {\frac{7}{3} \ne \operatorname{int} } \right)\,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,y = 2 \cdot 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{y^2}\left( { = {2^2} \cdot {3^2}} \right)\,\,\, = \,\,\,9\]
\[x = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{1^3}\left( { = 1} \right)\,\,\, = \,\,\,1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\]
\[x = {\text{prime}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\,\,\, = \,\,\,3 + 1 = 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\]
\[x = {\text{prime}}{{\text{s}}^{\,{\text{positive}}\,\,{\text{integers}}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\,\,\, = \,\,\,\,\left[ {({\text{posit}}\,{\text{mult}}\,\,{\text{of}}\,\,3) + 1} \right]\,\,\,{\text{product}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
