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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:

$B$

Community
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suomynonA
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2 Answers2

2

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.

Parcly Taxel
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  • How do you know that "The last natural number value this expression can possibly assume on the way down is 2"? – suomynonA Aug 14 '16 at 01:49
  • If $p>35$ the result of the expression must be in $(\frac32,2)$, which can never be an integer. We can thus ignore the interval $(35,\infty)$ for p. – Parcly Taxel Aug 14 '16 at 01:51
  • Sorry, can you explain how you know that If $p>35$ the result of the expression must be in $(\frac 32,2)$ – suomynonA Aug 14 '16 at 01:57
  • It might be best to plot a graph of the expression $y=\frac{3p+25}{2p-5}$ (you can use Desmos or Wolfram Alpha for this). The expression has the same shape as the $y=1/p$ graph, but its horizontal asymptote is $y=\frac32$, as the $3x$ and $2x$ terms come to dominate at large p. Then we note that $y(35)=2>\frac32$ and y is strictly decreasing from this point on. – Parcly Taxel Aug 14 '16 at 02:05
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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.

Milo Brandt
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