2.a) Find two natural numbers such that when divided, the quotient is 49, and the remainder is strictly greater than 4 and 294 less than the dividend.

b) Let $A=n^{2016}+(n+1)^{2016}+(n+2)^{2016}+\ldots+(n+9)^{2016}$, where $n \in \mathbb{N}$. Determine the remainder of the division of $A$ by 10.
2.a) Let $a$ and $b$ be the two numbers. Assume $a > b$.

The conditions from the problem can be written as:

$a = 49b + r \quad (1);$

$r > 4 \quad (2)$

$r = a - 294 \quad (3)$.

From equations (1) and (3), we get $a = 49b + a - 294$, which simplifies to $49b = 294$

$\Rightarrow b = 6$. Then $r \in \{0, 1, 2, 3, 4, 5\}$.

Since $r > 4 \Rightarrow r = 5$.

Thus, $a = 299$.

Therefore, the numbers sought are 299 and 6.

b) The remainder of the division of $A$ by 10 is the same as the last digit of $A$.

The numbers $n, n+1, n+2, \ldots, n+9$ have the last digits $0, 1, 2, \ldots, 9$, in some random order.

Thus, $u c(A) = u c\left(0^{2016} + 1^{2016} + 2^{2016} + \ldots + 9^{2016}\right) = $

$= u c(0 + 1 + 6 + 1 + 6 + 5 + 6 + 1 + 6 + 1) = 3$.

Therefore, the remainder of the division of $A$ by 10 is 3.