swerve wrote:If a, b, and c are consecutive positive integers and a < b < c, then what is the minimum possible value of
$$\frac{3^{2bc}}{3^{2ab}}?$$
A. 81
B. 2,187
C. 3,300
D. 6,561
E. 19,683
Source: EMPOWERgmat

$$\left( {a,b,c} \right) = \left( {a,a + 1,a + 2} \right)\,\,\,\,,\,\,\,a \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\,\left( * \right)$$
$$?\,\,\,\,:\,\,\,{\left( {{3^{\,2bc\, - \,2ab}}} \right)_{\,\min }}$$
$$?\,\,\,:\,\,\,{3^{2b\,\left( {c - a} \right)\,}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,\,\,{3^{2\,b\, \cdot \,2\,\,}}\,\mathop \ge \limits^{b\,\, \ge \,2} \,\,\,\,{3^{\,8}}\,\,\, = \,\,\,\,{3^{\,2}} \cdot {3^\,}^3 \cdot \,{3^{\,3}}\,\,\, = \,\,\,\,9 \cdot 27 \cdot 27$$
$$?\,\,\,:\,\,\,\,\left\{ \matrix{
\,{\rm{units}}\,\,{\rm{digit}}\,\, = \,\,1 \hfill \cr
\,{\rm{greater}}\,\,{\rm{than}}\,\,81 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( D \right)\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.