## Task B-1.3.

Determine the last two digits of the number $\underbrace{6^{6^{6} \cdot}}_{2019}$.
## First Solution.

Consider the powers $6^{n}, n \geqslant 2$.

$$
6^{2}=36,6^{3}=216,6^{4}=1296,6^{5}=\ldots 76,6^{6}=\ldots 56,6^{7}=\ldots 36,6^{8}=\ldots 16, \ldots \quad 1 \text { point }
$$

It is clear that the two-digit ending of the power $6^{n}, n \geqslant 2$ is one of the five numbers 36, 16, 96, 76, 56, which alternate in this order. That is, every fifth power has the same two-digit ending.

The power $6^{n}, n \geqslant 2$ has a specific two-digit ending depending on the remainder when the number $n-1$ is divided by 5.

If the remainder is 0, the ending is 56;

if the remainder is 1, the ending is 36;

if the remainder is 2, the ending is 16;

if the remainder is 3, the ending is 96;

if the remainder is 4, the ending is 76.

1 point

The number $6^{6^{6}}=6^{46656}$ has a two-digit ending of 56 because the number 46655 is divisible by 5.

Furthermore, the number $6^{6^{6^{6}}}=6^{6^{46656}}$ also has a two-digit ending of 56 because the exponent minus one is divisible by 5.

For the same reason, adding "another six" will result in each subsequent exponent having an ending of 56, so the number $\underbrace{6^{6^{6}}}_{2019}$ also has the last two digits 56.