, 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 $\text{2 6}$ 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$.