16. Esther and Frida are supposed to fill a rectangular array of 16 columns and 10 rows, with the numbers 1 to 160 . Esther chose to do it row-wise so that the first row is numbered $1,2, \ldots, 16$ and the second row is $17,18, \ldots, 32$ and so on. Frida chose to do it columnwise, so that her first column has $1,2, \ldots, 10$, and the second column has $11,12, \ldots, 20$ and so on. Comparing Esther's array with Frida's array, we notice that some numbers occupy the same position. Find the sum of the numbers in these positions.
\begin{tabular}{|c|c|c|cc|c|}
\hline 1 & 2 & 3 & $\cdots$ & $\cdots$ & 16 \\
\hline 17 & 18 & 19 & $\cdots$ & $\cdots$ & 32 \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline 145 & 146 & 147 & $\cdots$ & $\cdots$ & 160 \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|cc|c|}
\hline 1 & 11 & 21 & $\cdots$ & $\cdots$ & 151 \\
\hline 2 & 12 & 22 & $\cdots$ & $\cdots$ & 152 \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline 10 & 20 & 30 & $\cdots$ & $\cdots$ & 160 \\
\hline
\end{tabular}

Esther
Frida
16. Ans: 322. value $10(c-1)+r$ in Frida's array. So we need to solve
$$
16(r-1)+c=10(c-1)+r \Longrightarrow 5 r=3 c+2 .
$$

There are exactly four solutions
$$
(r, c)=\{(1,1),(4,6),(7,11),(10,16)\} .
$$