A problem for National Day

Let $c_k = k^2 - k + 1$ be the $k$ th central polygonal number. Then from $a_1$ to $a_{c_k}$ , there are $k$ central polygonal $a_n$ terms and $c_k - k$ non-central polygonal $a_n$ terms. Therefore,

$\begin{aligned} S_{c_k} & = 9(c_k-k) + k = 9c_k - 8k = 9(k^2-k+1)-8k = 9k^2-17k+9 \end{aligned}$

For $S_m = 2050$ , we have:

$\begin{aligned} 9k^2-17k + 9 & \approx 2050 \\ 9k^2-17k - 2041 & \approx 0 \\ \implies k & \approx 16.03 \end{aligned}$

Now, $S_{c_{16}} = 9(16^2) - 17(16)+9 = 2041$ . Therefore $S_{c_{16}+1} = 2041+9 = 2050$ , $\implies m = c_{16}+1 = 16^2-16+1+1 = \boxed{242}$ .