My first problem

$Let~ the ~polynomial ~be ~F(x) = x^3+ax^2 + bx + c \\F(0)=1~ \implies ~c=1......(1)\\ F(17)=5492 ~~ \implies ~ 17^3+17^2 a+ 17b =5491 =17*323\\\therefore ~17^2+17a+b=323 ~~ \implies ~17a + b = 34......(2)\\F(23)=13226 ~ \implies ~ 23^3+23^2a + 23b=13225=23*575\\\therefore ~23^2+23a+b=575 ~~ \implies ~23a + b = 46......(3)\\Solving~ (2) ~and~ (3) ~we ~get ~a=2~ and~ b=0 .....(4)\\ Leading~ coefficient~ is ~1~ and ~from~ (1) ~ and ~ (4)~ \\Answer = 1+a+b+c=1+2+0+1= \boxed{ \color{#20A900} {\huge {4} }}$

Let $F(x) = x^3+ax^2+bx+c$ . Then, we have:

$\begin{cases} F(0) = 0+0+0+c = 1 & \Rightarrow c = 1 & \\ F(17) = 4913+289a+17b +1=5492 & \Rightarrow 289a+17b = 578 &...(1)\\ F(23) = 12167+529a+23b+1=13226 & \Rightarrow 529a+23b = 1058 &...(2) \end{cases}$

$\begin{cases} Eq.1 \times 23: & 6647a+391b=13294 &...(1a) \\ Eq.2 \times 17: & 8993a+391b=17986 &...(2a) \end{cases}$

$Eq.2a-Eq.1a: 2346a +0 = 4692\quad \Rightarrow a = 2$

Substituting $\space a=2\space$ in $\space Eq.1: 578 +17b = 578 \quad \Rightarrow b = 0$

Therefore the sum of coefficients $= 1+a+b+c = 1+2+0+1 = \boxed{4}$