Generate this!

From $Q(5) = 154$ , we note that $n<4$ , since $a_j \ge 0$ and $5^4 > 154$ . Then we have:

$\begin{aligned} Q(x) & = a_0 + a_1x + a_2x^2 + a_3x^3 \\ \implies Q(5) & = a_0 + 5a_1 + 25a_2 + 125a_3 = 154 \end{aligned}$

The possible solutions are:

$\begin{cases} 4 + 5 \cdot 5 + 0 + 5^3 = 154 & \implies a_0 = 4, a_1 = 5, a_2 = 0, a_3 = 1 & \implies Q(1) = a_0 + a_1 + a_2 + a_3 = 10 \color{#D61F06}\ne 6 & \small \color{#D61F06} \text{Rejected} \\ 4 + 0 + 5^2 + 5^3 = 154 & \implies a_0 = 4, a_1 = 0, a_2 = 1, a_3 = 1 & \implies Q(1) = a_0 + a_1 + a_2 + a_3 \color{#3D99F6} = 6 & \small \color{#3D99F6} \text{Accepted} \end{cases}$

Therefore.

$\begin{aligned} Q(x) & = 4 + x^2 + x^6 \\ \implies Q(6) & = 4 + 36 + 216 = 256 \end{aligned}$

$\implies k = 256$ and $\sqrt {256} = 16$ which has a digital sum of $\boxed 7$ .

Since $Q(5)\equiv a_0\equiv 4\pmod{5}$ , we have $a_0=4$ . Since $Q(1)=6$ , we have $5^p+5^q=150$ for two positive integers $p$ and $q$ ; by inspection we find $p=2, q=3$ . Thus $Q(x)=4+x^2+x^3, Q(6)=256=16^2$ , and the answer is $1+6=\boxed{7}$ .