Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is the 6-digit positive integer abc000 divisible by 24?
1) The 3-digit integer abc is divisible by 8.
2) The 3-digit integer abc is divisible by 3.
$$\frac{{\left\langle {abc000} \right\rangle }}{{3 \cdot 8}}\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\frac{{\,1000 \cdot \left\langle {abc} \right\rangle }}{{3 \cdot 8}}\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\frac{{125 \cdot \left\langle {abc} \right\rangle }}{3}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,\,\,\frac{{\left\langle {abc} \right\rangle }}{3}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,\,\,\,\left( {a \ne {\text{0}}} \right)\,\,}$$
$$\left( 1 \right)\,\,\,{{\left\langle {abc} \right\rangle } \over 8}\,\, = \,\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left\langle {abc} \right\rangle = \left\langle {160} \right\rangle \,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\left[ {\sum\nolimits_{3\,\,{\rm{digits}}} { = 7\,\,{\rm{not}}\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,3} } \right] \hfill \cr
\,{\rm{Take}}\,\,\left\langle {abc} \right\rangle = \left\langle {168} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\left[ {\sum\nolimits_{3\,\,{\rm{digits}}} { = 15\,\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,3} } \right]\,\, \hfill \cr} \right.\,$$
$$\left( 2 \right)\,\,\,{{\left\langle {abc} \right\rangle } \over 3}\,\, = \,\,\,{\mathop{\rm int}} \,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
