## Task B-3.4.

Determine the last digit of the product of the first hundred natural numbers that give a remainder of 3 when divided by 5.
## Solution.

All natural numbers that give a remainder of 3 when divided by 5 can be written in the form $5k + 3, k \in \mathbb{N}_{0}$.

Let's write the product of the first hundred such numbers:

$$
\begin{aligned}
& 3 \cdot 8 \cdot 13 \cdot 18 \cdot 23 \cdot 28 \cdot 33 \cdots (5k + 3) \cdots 498 \\
& = (3 \cdot 8) \cdot (13 \cdot 18) \cdot (23 \cdot 28) \cdot (33 \cdot 38) \cdots ((5k - 2) \cdot (5k + 3)) \cdots (493 \cdot 498)
\end{aligned}
$$

Each of the 50 products in parentheses ends with the digit 4.

The last digit of our product is the last digit of the power $4^{50}$.

Since $4^{50} = 16^{25}$, and powers of 16 always end with the digit 6, the desired product ends with the digit 6.