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100 points for 49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A six-digit positive integer, ABCDEF, is divisible by all ##### This topic has 1 expert reply and 0 member replies ### Top Member ## A six-digit positive integer, ABCDEF, is divisible by all ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult 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 OA E Source: e-GMAT ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 Top Reply 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. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br • FREE GMAT Exam Know how you'd score today for0

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