Not sick of basick - 2

We can generate an equation:

$xp(x) = k(x-1)(x-2)(x-3)(x-4)(x-5) + 1$ .

Sub in $x=0$ we have:

$0 = -120k+1$

So $k = \frac{1}{120}$ and $xp(x) = \frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{120} + 1$ .

Sub in $x=6$ we get:

$6p(6) = \frac{5(4)(3)(2)(1)}{120} + 1 = 1+1 = 2$

$p(6) = \frac{2}{6} = \frac{1}{3}$ .

So $a+b =1+3 = \boxed{4}$

From the given information, we deduce that $p(x)$ is of the form:

$\begin{aligned} p(x) & = \frac{1}{24} (x-2)(x-3)(x-4)(x-5) - \frac{1}{12} (x-1)(x-3)(x-4)(x-5) \\ & \quad + \frac{1}{12} (x-1)(x-2)(x-4)(x-5) - \frac{1}{24} (x-1)(x-2)(x-3)(x-5) \\ & \quad \quad + \frac{1}{120} (x-1)(x-2)(x-3)(x-4) \end{aligned}$

$\begin{aligned} \Rightarrow p(6) & = \frac{1}{24} (4)(3)(2)(1) - \frac{1}{12} (5)(3)(2)(1) + \frac{1}{12} (5)(4)(2)(1) \\ & \quad - \frac{1}{24} (5)(4)(3)(1) + \frac{1}{120} (5)(4)(3)(2) \\ & = \frac{24}{24} - \frac{30}{12} + \frac{40}{12} - \frac{60}{24} + \frac{120}{120} \\ & = 1 - \frac{5}{2} + \frac{10}{3} - \frac{5}{2} + 1 \\ & = \dfrac{1}{3} \end{aligned}$

$\Rightarrow A + B = 1 + 3 = \boxed{4}$