Expression can be a positive integer if and only p is Question:

If p is a positive integer, then $\frac{3p+25}{2p-5}$ can be a positive integer, if and only if p is:
A) at least 3
B) at least 3 and no more than 35
C) no more than 35
D) equal to 35
E) equal to 3 or 35

I can eliminate D and E right off the bat, because if $p=5$ then the expression would be 8, a positive integer. That's pretty much it though; don't really know how to solve this.
Answer:

 $B$

 A: The function $\frac{3p+25}{2p-5}$ when treated as a continuous function is decreasing from the vertical asymptote at $p=\frac52$ towards $\frac32$ as $p\rightarrow\infty$. The last natural number value this expression can possibly assume on the way down is 2, and that is attained when $p=35$, which gives the upper bound in the answer. We now note that the expression is negative if $p<\frac52$, so p must be at least 3 (the lower bound) and the expression is fortunately a natural number (34) at this point. Hence the bounds on p are $3\le p\le35$, as the correct answer says.
A: The question you were given is extremely ambiguous - it is not at all clear what it means when an expression "can be" a positive integer. None of the given answers are exactly when it is a positive integer. A priori, anything "could" be an integer.
However, let us determine exactly when the expression $\frac{3p+25}{2p-5}$ is a positive integer. In order for it to be an integer at all, it must be that $$2p-5\,|\,3p+25$$
where $a|b$ means that $a$ divides $b$ - that is $\frac{b}a$ is a integer. Since the left hand side is odd, we can multiply the right hand side by two without changing divisibility. Thus, this is equivalent to
$$2p -5\,|\,6p+50.$$
We may always add a multiple of the left hand side to the right hand side without changing its truth. Adding $-6p + 15$ to both sides gives
$$2p - 5\,|\,65.$$
Thus, this is satisfied exactly when $2p-5$ is a factor of $65$. The only odd factors of $65$ are $\pm1,\pm5,\,\pm13,\,\pm65$, thus the only possible positive $p$ are $1,\,3,\,5,\,9,\,35$ and $1$ yields that the fraction is a negative integer. Thus, the only possible $p$ are $3,\,5,\,9,$ and $35$. The most restrictive answer which allows this is B.
