Problem Solving

vaibhav101 wrote:When 38 is subtracted from a number N, the result obtained has 341 as the sum of the digits. If N is divisible by 10^r but N is not divisible by 10^(r+1), what is the value of r ?
A 35
B 36
C 37
D 38
E 39

\[\frac{N}{{{{10}^{r\, \geqslant 1}}}} = \operatorname{int} \,\,\,\, \Rightarrow \,\,\,N\,\,{\text{has}}\,\,{\text{at}}\,\,{\text{least}}\,\,{\text{r}}\,\,{\text{zeros}}\,\,{\text{as}}\,\,{\text{last}}\,\,{\text{digits}}\]
\[\frac{N}{{{{10}^{r + 1}}}} \ne \operatorname{int} \,\,\,\, \Rightarrow \,\,\,N\,\,{\text{does}}\,\,{\text{NOT}}\,\,{\text{have}}\,\,{\text{more}}\,\,{\text{than}}\,\,{\text{r}}\,\,{\text{zeros}}\,\,{\text{as}}\,\,{\text{last}}\,\,{\text{digits}}\]
Let´s observe a certain pattern:
\[{10^2} - 38 = 62\,\,\,\, \to \,\,\,\,\sum\nolimits_{digits} { = 6 + 2} \]
\[{10^3} - 38 = 962\,\,\,\, \to \,\,\,\,\sum\nolimits_{digits} { = 9 + 6 + 2} \]
\[{10^4} - 38 = 9962\,\,\,\, \to \,\,\,\,\sum\nolimits_{digits} { = 2 \cdot 9 + 6 + 2} \]
\[{10^5} - 38 = 99962\,\,\,\, \to \,\,\,\,\sum\nolimits_{digits} { = 3 \cdot 9 + 6 + 2} \]

By "vulgar induction", we know that:
\[{10^M} - 38 = 9 \ldots 962\,\,\,\, \to \,\,\,\,\sum\nolimits_{digits} { = \left( {M - 2} \right) \cdot 9 + 6 + 2} \]
\[\left( {M - 2} \right) \cdot 9 + 6 + 2 = 341\,\,\, \Rightarrow \,\,\,M = 39\]
We have proved that N=10^39 satisfies the question stem, therefore the answer must be 39. (*)

Regards,
fskilnik.

(*) P.S.: N does not need to be, a priori, a "pure" power of 10. My solution "searches" only this particular case, hence it is incomplete.
(For example: 2*(10^39) has also exactly 39 zeros as final digits, although 2*(10^39) -38 does not have the sum of digits required, of course.)
I have decided to avoid the "general case" because, I believe, it is out-of-GMAT´s scope.