Problem 248

Since $A$, $B$ and $C$ are collinear, segments $AB$ and $AC$ (and $BC$) have the same slope.

The slope of segment $AB$ is $\dfrac{5-2}{2-1} = 3$. The slope of
segment $AC$ is $\dfrac{c-2}{3-1} = \dfrac{c-2}{2}$.

Therefore, we have that $3 =
$\dfrac{c-2}{2}$$. Solving for$c$gives us that$c=8$.
Since segment $DF$ has slope
$\dfrac{7-0}{0-14} = - \dfrac{1}{2}$
and $y$-intercept $7$, the equation of the line

through points $D$ and $F$ is $y=-\dfrac{1}{2}x + 7$.

The coordinates $(x,y)$ of $E$ must be positive integers and $E$ must lie on the line $y=-\dfrac{1}{2}x + 7$. When $x \geq 14$ we have that $y \leq 0$, and when $x < 14$ we have that $y>0$. Therefore $x$ and $y$ are both positive exactly when $0<x<14$. Also, $y$ is an integer exactly when $x$ is even.

Therefore the number of possible points $E$ is the number of integers $0 < x < 14$ such that $x$ is even. There are $6$ such integers which give the following
$6$ possibilities for $E$: $(2,6)$, $(4,5)$, $(6,4)$, $(8,3)$, $(10,2)$, and $(12,1)$.
The slope of segment $PQ$ is
$\dfrac{n-4-12}{6-15}=\dfrac{n-16}{-9}$.

The slope of segment $PR$ is $\dfrac{n-12}{18-15}=\dfrac{n-12}{3}$.

Therefore, points $P$, $Q$ and $R$ are collinear exactly when $\dfrac{n-16}{-9}=\dfrac{n-12}{3}$. This is
equivalent to $3n-48 = -9n+108$,
which is equivalent to $12n=156$.

Therefore, the value of $n$ that
makes $P$, $Q$ and $R$ collinear is $n=\dfrac{156}{12}=13$.