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A six-digit positive integer, ABCDEF, is divisible by all

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A six-digit positive integer, ABCDEF, is divisible by all

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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

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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

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