We call the "last digit of a number" as the digit farthest to the right. For example, the last digit of 2014 is the digit 4.

a) What is the last digit of $11^{11}$?

b) What is the last digit of $9^{9}$? And what is the last digit of $9219^{9219}$?

c) What is the last digit of $2014^{2014}$?
Solution

a) Observe that

$$
11^{11}=\underbrace{11 \times 11 \times 11 \times 11 \times \cdots \times 11}_{11 \text { times }}
$$

Each time we multiply two numbers that have the digit 1 as the rightmost digit, we get a number that also has the digit 1 as the rightmost digit. For example, $11 \times 11=121$. Therefore, repeating the process, we will find that $11^{11}$ has the digit 1 as the rightmost digit.

b) Observe that

$$
9^{9}=\underbrace{9 \times 9 \times \cdots \times 9}_{9 \text { times }}
$$

In this case, when we perform the first multiplication, we get $9 \times 9=81$, which ends in 1. When we perform the next multiplication, we get $81 \times 9=729$, which ends in 9. Performing the next multiplication, we get a number that ends in 1 again. Then another number, now ending in 9. Thus, we have a pattern! Since we start with 9 and perform eight multiplications, we will end up with a number that ends in 9.

For the number $9219^{9219}$, we perform the same analysis as before. Since we start with a number that ends in 9 and perform 9218 multiplications, we conclude that the final result will be a number that ends in 9.

c) In this case, let's observe the pattern generated again. Note that $4 \times 4=16$. Multiplying the last digit (which in this case is 6) by 4, we get $6 \times 4=24$. Thus, the last digit becomes 4 again. In other words, we get a pattern where the last digit alternates between 4 and 6. Since

$$
2014^{2014}=\underbrace{2014 \times 2014 \times \cdots \times 2014}_{2014 \text { times }}
$$

we start with a 2014 and perform 2013 multiplications. Therefore, the rightmost digit of the result will be the digit 6.