, 2025

Consider the function $g(x)=59x$. Observe that $g(1)=59$, $g(2)=118$, and $g(3)=177$, from which it follows that
$f(1)=g(1)$, $f(2)=g(2)$, and $f(3)=g(3)$.

Now define a polynomial $h(x)=f(x)-g(x)$ so that, by construction,
$x=1$, $x=2$, and $x=3$ are roots of $h(x)$.

Algebraically, we also have that $$h(x) = f(x)-g(x) = x^4+px^3+qx^2+rx+s - 59x = x^4 + px^3 + qx^2 + (r-59)x+s$$ but the important thing to notice here is that $h(x)$ is a degree four polynomial with a
leading coefficient of $1$.

Since we already know that $x=1$,
$x=2$, and $x=3$ are roots of $h(x)$, we conclude that $$h(x)=(x-1)(x-2)(x-3)k(x)$$ for some
polynomial $k(x)$. However, because
$h(x)$ has degree $4$ and a leading coefficient of $1$, it must be true that $k(x)$ has degree $1$ and a leading coefficient of $1$.

Thus, $h(x) = (x-1)(x-2)(x-3)(x-a)$
for some real number $a$.

We now evaluate $h(x)$ at $x=9$ and $x=-5$ to get $$\begin{align*} h(9) &= (9-1)(9-2)(9-3)(9-a) = 336(9-a) \\ h(-5) &= (-5-1)(-5-2)(-5-3)(-5-a) = 336(5+a)\end{align*}$$
Now using that $h(x) = f(x)-g(x)$ or
$f(x) = h(x)+g(x)$, we can determine
the value of $f(9)+f(-5)$ as follows:
$$\begin{align*} f(9) + f(-5) &= h(9)+g(9) + h(-5)+g(-5) \\ &= 336(9-a) + 9(59) + 336(5+a) -5(59) \\ &= 336(9)-336a+9(59)+336(5)+336a-5(59) \\ &= 336(9+5)+59(9-5) \\ &= 336(14)+59(4) \\ &= 4940\end{align*}$$ Therefore, $T=4940$, so the answer is $4+9+4+0=17$.