Wierd Remainder

We separate the expression into two parts. Notice that

$16+13k=13+3+13k=13(k+1)+3\equiv 3 \pmod{13}$ .

While, $140-13i=10(13)+10-13i=13(10-i)+10\equiv 10 \pmod{13}$ .

Thus,

$\displaystyle\prod^{30}_{k=0} (16+13k) \equiv 3^{31} \equiv (3^3)^{10} \times 3 \equiv 1^{10}\times 3 \equiv 3 \pmod{13}$

and

$\displaystyle\sum^{10}_{i=0} (140-13i) \equiv 10\times 11\equiv 110\equiv 6 \pmod{13}$

So, the remainder is just simply $3+6=\boxed{9}$

It is noted that the product can be rewritten as follows:

$P = \prod_{k=0}^{30} {(16+13k)} = \prod_{k=1}^{31} {(3+13k)}$

Since $3+13k \equiv 3 \pmod {13}$

$\Rightarrow P \mod {13} = 3^{31} \mod {13}$

It can be shown that $3^n \mod {13} = 3(n \mod {3}) \Rightarrow 3^{31} \mod {13} = 3(31 \mod {3} ) = 3$

Similarly, $S = \sum _{i=0} ^{10} { (140 - 13i) } = \sum _{i=0} ^{10} { (10 + 13i) }$

$\Rightarrow S \mod {13} = 110 \mod {13} = 6$

Therefore, $S+P \equiv (3+6) \pmod {13} \equiv \boxed{9} \pmod {13}$