alanforde800Maximus wrote:After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?
A. None
B. 1
C. 2
D. 8
E. 9
$$\left. \matrix{
A = \,\,\underbrace {\underline {} \,\,\underline {} \,\,\underline {} \,\, \ldots \,\,\underline {} }_{n\,\,{\rm{digits}}}\,\,\,\, \ge 1\,\,{\mathop{\rm int}} \hfill \cr
A \cdot \left( {n + 2} \right) = \underbrace {\underline {n + 1} \,\,\underline {n + 1} \,\,\underline {n + 1} \,\, \ldots \,\,\underline {n + 1} }_{n + 1\,\,{\rm{digits}}}\,\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,?\,\,\,\, = \,\,\,\,\# \,\,A\,\,\,{\rm{possible}}$$
This is a typical organized manual work technique exercise!
$$\left\{ \matrix{
n = 1\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\, \cdot \,\,\,\left( {1 + 2} \right) = \underline 2 \,\,\underline 2 \,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{22} \over 3} \ne {\mathop{\rm int}} } \right) \hfill \cr
n = 2\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {2 + 2} \right) = \underline 3 \,\,\underline 3 \,\,\underline 3 \,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{{\rm{odd}}} \over {{\rm{even}}}} \ne {\mathop{\rm int}} } \right) \hfill \cr
n = 3\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {3 + 2} \right) = \underline 4 \,\,\underline 4 \,\,\underline 4 \,\,\underline 4 \,\,\,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{4444} \over 5} \ne {\mathop{\rm int}} } \right) \hfill \cr
n = 4\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr
n = 5\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {5 + 2} \right) = \underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\,\,\,\,\,,\,\,\,\,\underline {{\rm{viable}}} \,\,{\rm{solution}}\,\,\,\,\,\left( {{{666666} \over 7} = 95238} \right) \hfill \cr
n = 6\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr
n = 7\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {7 + 2} \right) = \underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{88888888} \over 9} \ne {\mathop{\rm int}} } \right) \hfill \cr
n = 8\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr
n = 9\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {9 + 2} \right) = \underline {10} \,\,\underline {10} \,\,\underline {10} \,\, \ldots \,\,\underline {10} \,\,\,????\,\,\,\,{\rm{impossible}}\, \hfill \cr
n \ge 10\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 9} \right) \hfill \cr} \right.$$
The correct answer is therefore (B). (This is all VERY fast, although hard to type!)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
Featured Course
