## Task B-1.5.

Determine the last two digits of the number $7^{2016^{2017}}$.
## Solution.

The number $7^{2016^{2017}}$ can be written as follows:

$$
7^{2016^{2017}}=7^{2016 \cdot 2016 \cdot 2016 \cdots \cdots \cdot 2016}
$$

The exponent of the given number is divisible by 8 because 2016 is divisible by 8, so we have:

$$
7^{2016^{2017}}=7^{8 x}=\left(7^{4}\right)^{2 x}
$$

where $x$ is some integer. Further, we have

$$
7^{2016^{2017}}=\left(7^{4}\right)^{2 x}=(2400+1)^{2 x}=\left(2400^{2}+4800+1\right)^{x}=(100 y+1)^{x} \quad 1 \text { point }
$$

where $y$ is also some integer.

The last two digits of the given number are 01.

Note: Students can observe the powers of 7 and conclude that they always end in $7,9,3$ and 1, and since the exponent of the given number is divisible by 4, it must have the last digit 1. (3 points)

Since $7^{2016^{2017}}=7^{4 k}=2401 \cdots 2401 \cdot 2401$, the product will certainly end with 0 in the second-to-last position (which is worth the remaining 3 points), and this needs to be justified as in the first solution. Accept also if the student has justified it using part of the written multiplication with the last number 2401.