BTGmoderatorDC wrote:If a, b, and c are positive integers, what is the remainder when a - b is divided by 6?
(1) a = c^3
(2) b = (c - 2)^3
Source: Manhattan Prep
$$a,b,c\,\, \ge 1\,\,\,{\rm{ints}}\,\,\,\,\left( * \right)$$
$$a - b = 6Q + r\,\,,\,\,\,\left\{ \matrix{
\,Q,r\,\,\,{\mathop{\rm int}} {\rm{s}} \hfill \cr
\,0 \le r \le 5 \hfill \cr} \right.$$
$$? = r$$
$$\left( 1 \right)\,\,a = {c^3}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,1} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 0 \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {8,1,2} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 1 \hfill \cr} \right.$$
$$\left( 2 \right)\,\,b = {\left( {c - 2} \right)^3}\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 0 \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {2,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 1 \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,a - b = {c^3} - {\left( {c - 2} \right)^3} = \ldots = \underbrace {6{c^2} - 12c + 6}_{6Q\,\,,\,\,Q = {c^2} - 2c + 1} + 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 2$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
