BTGmoderatorLU wrote:Source: Manhattan Prep
If a, b, and c are positive integers such that a < b < c, is a% of b% of c an integer?
1) b = (a/100)^-1
2) c = 100^b
$$1 \le a < b < c\,\,\,{\rm{ints}}\,\,\,\left( * \right)$$
$$\left[ {{a \over {100}}\left( {{b \over {100}}\left( c \right)} \right) = } \right]\,\,\,\,\,{{abc} \over {{{10}^4}}}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}} $$
$$\left( 1 \right)\,\,b = {{100} \over a}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,100,1000} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,100,101} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.$$
$$\left( 2 \right)\,\,c = {\left( {{{10}^2}} \right)^b}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,c = {\left( {{{10}^2}} \right)^{ \ge \,\,2}} = {10^{ \ge \,4}}$$
$${{abc} \over {{{10}^4}}} = {{ab \cdot {{10}^{ \ge \,4}}} \over {{{10}^4}}}\,\,\mathop = \limits^{\left( * \right)} \,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
