CEMC Euclid, 2026

Using the prime factorization of each of the factors of the
numerator, we see that $$\begin{align*} 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 & = 3^2 \cdot 2^3 \cdot 7 \cdot (2 \cdot 3) \cdot 5 \cdot 2^2 \cdot 3 \\ & = 2^6 \cdot 3^4 \cdot 5 \cdot 7\end{align*}$$ To find the
smallest positive integer $n$ for
which $\dfrac{2^6 \cdot 3^4 \cdot 5 \cdot 7}{n}$ is a perfect cube, we look for the minimal set of prime
divisors that we can remove (that is, divide out) from the numerator so
that the number of times that each remaining prime occurs is a multiple
of $3$. This is because one way of
characterizing a perfect cube is that each of its prime factors occurs
in groups of $3$.

To do this, we need to remove at least $1$ factor of $3$, at least $1$ factor of $5$, and at least $1$ factor of $5$. This means that $n \geq 3 \cdot 5 \cdot 7$.

If $n = 3\cdot 5 \cdot 7 = 105$, then
$$\dfrac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{n} = 2^6 \cdot 3^3 = (2^2 \cdot 3)^3$$ Since
$n \geq 105$ and $n=105$ gives a perfect cube, then the
smallest possible $n$ is $n = 105$.
Since $3^{a+b} = 27$ and $3^3 = 27$, then $a + b = 3$.

Adding $a+b=3$ to the equation $a - b = -5$, we obtain $2a = -2$ and so $a = -1$.

Since $b = 3 - a$, then $b = 4$ and so $(a, b) = (-1, 4)$.
Since $P(10, 0)$ lies on the
parabola with equation $y = -x^2 + 7x + c$, then $0 = -100 + 70 + c$
and so $c = 30$.

Thus, the parabola has equation $y = -x^2 + 7x + 30$ which can be factored to obtain $y = -(x-10)(x+3)$.

Since $Q$ is the other $x$-intercept of the parabola, then $Q$ has coordinates $(-3, 0)$.

Since $R$ is the point where the
parabola crosses the $y$-axis, we set
$x = 0$ and obtain $y = 30$.

Thus, we want to find the area of the triangle with vertices $P(10, 0)$, $Q(-3, 0)$ and $R(0, 30)$.

We note that $PQ$ is horizontal so
can be treated as the base of the triangle. Also, $PQ = 10 - (-3) = 13$.

Point $R$ is $30$ units above $PQ$, so the height of the triangle
relative to base $PQ$ is $30$.

Therefore, the area of $\triangle PQR$ is $\frac{1}{2} \cdot 13 \cdot 30 = 195$.