BTGmoderatorDC wrote:A six-digit positive integer, ABCDEF, is divisible by all the integers from 2 to 6, both inclusive. How many such numbers are possible, if all the digits of the number are distinct and less than 7?
A. 24
B. 48
C. 72
D. 96
E. 120
Source: e-GMAT
Very nice problem, but absolutely not GMAT-time-like!
$${{\left\langle {ABCDEF} \right\rangle } \over {LCM\left( {2,3,4,5,6} \right)}} = {\mathop{\rm int}} $$
$$\left\langle {ABCDEF} \right\rangle \,\,\,\,{\rm{with}}\,\,\,\left\{ \matrix{
\,{\rm{A}} \in \left\{ {1,2, \ldots ,6} \right\} \hfill \cr
\,{\rm{B,C,}} \ldots {\rm{,F}}\,\, \in \left\{ {0,1,2, \ldots ,6} \right\} \hfill \cr} \right.\,\,\,\,\,\,{\rm{all}}\,\,{\rm{distinct}}\,{\rm{digits}}\,\,\,\,\,\,\left( * \right)$$
$$?\,\,\,:\,\,\,\,\# \,\,\,\left\langle {ABCDEF} \right\rangle \,\,\,{\rm{possible}}$$
$$\left( 1 \right)\,\,\,{{\left\langle {ABCDEF} \right\rangle } \over {2 \cdot 5}} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,F = 0\,\,\,\,\,\,\,\left( { \Rightarrow \,\,\sum {\,{\rm{other}}\,\,{\rm{digits}} = 21} } \right)$$
$$\left( 2 \right)\,\,\,{{\left\langle {ABCDE0} \right\rangle } \over 4} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{{\left\langle {E0} \right\rangle } \over 4} = {\mathop{\rm int}} \,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,E \in \left\{ {2,4,6} \right\}$$
$$\eqalign{
& \left( {3{\rm{a}}} \right)\,\,\left( {E,F} \right) = \left( {2,0} \right)\,\,\, \Rightarrow \,\,\,\,\sum {{\rm{other}}\,\,{\rm{digits}} = 19} \,\,{\rm{and}}\,\,\left\{ {A,B,C,D} \right\} \subset \left\{ {1,3,4,5,6} \right\}\,\,{\rm{with}}\,\,{\rm{remainder}}\,\,1\,\,{\rm{when}}\,\,{\rm{divided}}\,\,{\rm{by}}\,\,3 \cr
& \left( {3{\rm{b}}} \right)\,\,\left( {E,F} \right) = \left( {4,0} \right)\,\,\, \Rightarrow \,\,\,\,\sum {{\rm{other}}\,\,{\rm{digits}} = 17} \,\,{\rm{and}}\,\,\left\{ {A,B,C,D} \right\} \subset \left\{ {1,2,3,5,6} \right\}\,\,{\rm{with}}\,\,{\rm{remainder}}\,\,2\,\,{\rm{when}}\,\,{\rm{divided}}\,\,{\rm{by}}\,\,3 \cr
& \left( {3{\rm{c}}} \right)\,\,\left( {E,F} \right) = \left( {6,0} \right)\,\,\,\, \Rightarrow \,\,\,\sum {{\rm{other}}\,\,{\rm{digits}} = 15} \,\,{\rm{and}}\,\,\left\{ {A,B,C,D} \right\} \subset \left\{ {1,2,3,4,5} \right\}\,{\rm{ with}}\,\,{\rm{remainder}}\,\,0\,\,{\rm{when}}\,\,{\rm{divided}}\,\,{\rm{by}}\,\,3 \cr} $$
$${{\left\langle {ABCDE0} \right\rangle } \over 3} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{{A + B + C + D + E} \over 3} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left( {3{\rm{a}}} \right)\,\,\,\, \Rightarrow \left\{ {A,B,C,D} \right\} \subset \left\{ {1,4,5,6} \right\}\,\,\,{\rm{or}}\,\,\,\,\left\{ {A,B,C,D} \right\} \subset \left\{ {1,3,4,5} \right\}\, \hfill \cr
\,\left( {3{\rm{b}}} \right)\,\,\,\, \Rightarrow \left\{ {A,B,C,D} \right\} \subset \left\{ {1,2,5,6} \right\}\,\,\,{\rm{or}}\,\,\,\,\left\{ {A,B,C,D} \right\} \subset \left\{ {1,2,3,5} \right\}\, \hfill \cr
\,\left( {3{\rm{c}}} \right)\,\,\,\, \Rightarrow \left\{ {A,B,C,D} \right\} \subset \left\{ {1,2,4,5} \right\}\,\, \hfill \cr} \right.$$
$$? = 5 \cdot 4! = 5! = 120\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
