8. What are the last two digits of $7^{2018}$ ?
A 07
B 49
C 43
D 01
E 18
Solution
B
Commentary
Clearly, we cannot answer this question by fully evaluating $7^{2018}$ and then looking to see which are its last two digits.
Instead, we make use of the fact that the last two digits of a product $a \times b$ is determined just by the last two digits of $a$ and $b$, and then we look for a pattern.

It is convenient to introduce some notation for the last two digits of an integer. There is no standard notation for this (but see 8.1, below). For the purpose of this question we use the notation $[n]$ for the number consisting of the last two digits of the integer $n$. For example, $[12345]=45$.
Using this notation we can express the fact we use by the equation
$$
[m \times n]=[[m] \times[n]] .
$$

We then have
$$
\begin{array}{l}
{\left[7^{1}\right]=[7]=07} \\
{\left[7^{2}\right]=[49]=49} \\
{\left[7^{3}\right]=\left[7^{2} \times 7\right]=\left[\left[7^{2}\right] \times[7]\right]=[49 \times 7]=[343]=43} \\
{\left[7^{4}\right]=\left[7^{3} \times 7\right]=\left[\left[7^{3}\right] \times[7]\right]=[43 \times 7]=[301]=01} \\
{\left[7^{5}\right]=\left[7^{4} \times 7\right]=\left[\left[7^{4}\right] \times[7]\right]=[01 \times 7]=[7]=07 .}
\end{array}
$$

Each term in the sequence giving the last two digits of $7^{n}$ for $n=1,2,3 \ldots$ depends only on the previous term. Hence as 07 has reoccurred we can deduce that the sequence consisting of the values of $\left[7^{n}\right]$ is made up of repeating cycle of length 4 , and so is
$$
07, \quad 49, \quad 43, \quad 01, \quad 07, \quad 49, \quad 43, \quad 01, \quad \ldots
$$

Because $2018=504 \times 4+2$ it follows that $\left[7^{2018}\right]$ is the second pair of digits in this cycle, namely, 49 .