Question 1209697
Let the roots of $f(x)$ be $-r_1, -r_2, -r_3$, where $r_1, r_2, r_3$ are positive integers.
Then $f(x) = (x+r_1)(x+r_2)(x+r_3) = x^3 + (r_1+r_2+r_3)x^2 + (r_1r_2 + r_1r_3 + r_2r_3)x + r_1r_2r_3$.
Comparing this with $f(x) = x^3 + ax^2 + bx + c$, we have:
$a = r_1+r_2+r_3$
$b = r_1r_2 + r_1r_3 + r_2r_3$
$c = r_1r_2r_3$

We are given $a+b+c = 10$, so
$r_1+r_2+r_3 + r_1r_2 + r_1r_3 + r_2r_3 + r_1r_2r_3 = 10$.
Adding 1 to both sides, we get
$1 + r_1+r_2+r_3 + r_1r_2 + r_1r_3 + r_2r_3 + r_1r_2r_3 = 11$
$(1+r_1)(1+r_2)(1+r_3) = 11$
Since $r_1, r_2, r_3$ are positive integers, $1+r_1, 1+r_2, 1+r_3$ must be integers greater than 1.
Since 11 is a prime number, the only possible factorization is $11 = 11 \cdot 1 \cdot 1$.
Thus, we must have $1+r_1 = 11$, $1+r_2 = 1$, and $1+r_3 = 1$. This means $r_1 = 10$, $r_2 = 0$, and $r_3 = 0$. However, the roots must be positive integers, so $r_2$ and $r_3$ cannot be 0.

Alternatively, we can have $1+r_1 = 11$, which means $r_1 = 10$. Then $(1+r_2)(1+r_3) = 1$, which is not possible since $r_2$ and $r_3$ are positive integers.

If we let $1+r_1=11$, $1+r_2=1$ and $1+r_3=1$, then $r_1=10$, $r_2=0$ and $r_3=0$, which are not all positive integers.
However, if we consider the case where two of the integers are 1, then we have
$(1+r_1)(1+r_2)(1+r_3) = 11$.
Let $r_2 = r_3 = 0$. Then $1+r_1 = 11$, so $r_1 = 10$.
$a = 10+0+0 = 10$
$b = 0+0+0 = 0$
$c = 0$
$a+b+c = 10+0+0 = 10$.
So $c = 0$.

Final Answer: The final answer is $\boxed{0}$