Problem Solving

vaibhav101 wrote:For how many integers, between 1 and 2009, is the improper fraction (n^2+4)/(n+5) not in its simplest form?
A 67
B 69
C 200
D 57
E 29

Dear vaibhav101,

I believe this problem is out-of-GMAT´s scope. In respect to the interest of this community, I will not answer other problems of yours that, in my opinion, are in similar situation.
Suggestion: post here, for instance: https://math.stackexchange.com/
Thank you very much for your understanding!

\[{n^2} + 4 = {n^2} - 25 + 29 = \left( {n - 5} \right)\left( {n + 5} \right) + 29\,\,\, \Rightarrow \,\,\,\frac{{{n^2} + 4}}{{n + 5}} = \underbrace {n - 5}_{\operatorname{int} } + \frac{{29}}{{n + 5}}\]
\[\frac{{{n^2} + 4}}{{n + 5}}\,\,{\text{not}}\,\,{\text{irreducible}}\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,n + 5\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{multiple}}\,\,{\text{of}}\,\,29\]
(*) For the interested readers: try (say) 8/7 , 13/8 and then 14/8, 15/8, 16/8 to "feel" what is involved here...

The continuation is absolutely inside GMAT´s scope, therefore relevant to our purposes!
\[1 < n < 2009\,\,\, \Leftrightarrow \,\,\,6 < n + 5 = 29M < 2014\,\,\,\,\,\,\left( {M\,\,\operatorname{int} } \right)\]
\[29 \leqslant 29M \leqslant 2001\,\,\, \Leftrightarrow \,\,\,1 \leqslant M \leqslant 69\,\,\,\,\]
\[? = 69\]

Regards,
fskilnik.