Problem Solving

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

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.