Junior Balkan Mathematical Olympiad

Solution:
Grouping the elements of the product by ten we get:
$$\begin{aligned} & (30k+1)(30k+4)(30k+7)(30k+10)(30k+13)(30k+16) \\ & (30k+19)(30k+22)(30k+25)(30k+28) = \\ & = (30k+1)(15k+2)(30k+7)(120k+40)(30k+13)(15k+8) \\ & (30k+19)(15k+11)(120k+100)(15k+14) \end{aligned}$$
(We divide all even numbers not divisible by five, by two and multiply all numbers divisible by five with four.)
We denote $P_k = (30k+1)(15k+2)(30k+7)(30k+13)(15k+8)(30k+19)(15k+11)(15k+14)$. For all the numbers not divisible by five, only the last digit affects the solution, since the power of two in the numbers divisible by five is greater than the power of five. Considering this, for even $k$, $P_k$ ends with the same digit as $1 \cdot 2 \cdot 7 \cdot 3 \cdot 8 \cdot 9 \cdot 1 \cdot 4$, i.e. six and for odd $k$, $P_k$ ends with the same digit as $1 \cdot 7 \cdot 7 \cdot 3 \cdot 3 \cdot 9 \cdot 6 \cdot 9$, i.e. six. Thus $P_0 P_1 \ldots P_{66}$ ends with six. If we remove one zero from the end of all numbers divisible with five, we get that the last nonzero digit of the given product is the same as the one from $6 \cdot 2011 \cdot 2014 \cdot 4 \cdot 10 \cdot 16 \cdot \ldots \cdot 796 \cdot 802$. Considering that $4 \cdot 6 \cdot 2 \cdot 8$ ends with four and removing one zero from every fifth number we get that the last nonzero digit is the same as in $4 \cdot 4^{26} \cdot 784 \cdot 796 \cdot 802 \cdot 1 \cdot 4 \cdot \ldots \cdot 76 \cdot 79$. Repeating the process we did for the starting sequence we conclude that the last nonzero number will be the same as in $2 \cdot 6 \cdot 6 \cdot 40 \cdot 100 \cdot 160 \cdot 220 \cdot 280 \cdot 61 \cdot 32 \cdot 67 \cdot 73 \cdot 38 \cdot 79$, which is two.