## Task 1 - 050731

What digits does the product end with?

$$
z=345926476^{3} \cdot 125399676^{2} \cdot 2100933776^{3}
$$
All factors of $z$ are of the form 100a+76. Since

$$
(100 a+76)(100 b+76)=10000 a b+100(76 a+76 b)+76^{2}
$$

where $a$ and $b$ are natural numbers, only the last two digits of $76^{2}$ are decisive for the last two digits of $z$. Since $76^{2}=5776$, $z$ ends in 76.