Tomi's Challenges - #22

Denote the left-hand side as $f(x)$ . Factoring, $f(x) = (x-4)(x-5)(x-6) \dots (x-11)(x-12)(x-13)$ .

By inspection, $\boxed{18}$ is an $x$ -value that works. But note that $f(x)$ is a strictly increasing function when $x \ge 14$ , hence it is the only value that works.

$\begin{aligned} (x^2-9x+20)\blue{(x^2-13x+42)}\red{(x^2-17x+72)}(x^2-21x+110)\blue{(x^2-25x+156)} & =\frac{14!}{4!} \\ (x-4)(x-5) \blue{(x-6)(x-7)} \red{(x-8)(x-9)} (x-10)(x-11) \blue{(x-12)(x-13)} & = 5 \times 6 \times 7 \cdots \times 14 \\ \implies x & = - 1, 18 \end{aligned}$

Therefore the sum of natural numbers $x \ge 14$ satisfying the equation above is $\boxed {18}$ .

Divide both sides by $10!$ to generate a binomial coefficient on the right-hand side. The quadratics on the left-hand side can be factored into $\frac{(x-4)(x-5)\cdots(x-13)}{10!}=\binom{x-4}{10}\overset{!}{=}\binom{14}{10} \quad\Rightarrow\quad x=\boxed{18}$ For $x\geq 14$ the left-hand side is strictly increasing, so it is the only solution.

The first step is to factor all the expressions in the factors. $x^2-9x+20=x^2-5x-4x+20=x(x-5)-4(x-5)=(x-4)(x-5)$ . Notice that $42=6\cdot 7, 13=6+7$ , $72=8\cdot 9, 17=8+9$ , $110=10\cdot 11, 21=10+11$ , $156=12\cdot 13, 25=12+13$ , which suggests that the other factors follow a similar pattern. Overall we have:

$(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)(x-11)(x-12)(x-13)=\frac{14!}{4!}$ $4!(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)(x-11)(x-12)(x-13)=14!$ (since $x\geq 14$ ) $4!(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)(x-11)(x-12)(x-13)(x-14)!=14!(x-14)!$ $4!(x-4)!=14!(x-14)!$ $\frac{4!(x-4)!}{x!}=\frac{14!(x-14)!}{x!}$ $\frac{x!}{4!(x-4)!}=\frac{x!}{14!(x-14)!}$ $\binom{x}{4}=\binom{x}{14}$

Therefore $x=14+4=18$ .