[Periodicity and Non-periodicity]  [Decimal Number System]

What is the last digit of the number $1989^{1989}$? And what are the last digits of the numbers $1989^{1992}$, $1992^{1989}$, $1992^{1992}$?
Try to determine the last digits of the numbers $9^{1989}, 9^{1992}, 2^{1989}, 2^{1992}$.

## Solution

Since we are only interested in the last digits of the results, it is sufficient to determine the last digits of the numbers $9^{1989}, 9^{1992}, 2^{1989}$, and $2^{1992}$.

The number 9, when raised to a power, gives two possible last digits - 9 (if the power is odd) and 1 (if the power is even). This means that $9^{1989}$ has the last digit 9, and $9^{1992}$ has the last digit 1.

The number 2, when raised to a power, can give the following last digits: 2, 4, 8, 6. If the exponent divided by 4 leaves a remainder of 1 - the last digit will be 2; if the remainder is 2 - the last digit will be 4; if the remainder is 3 - the last digit will be 8; if there is no remainder - the last digit will be 6. This means that $2^{1989}$ has the last digit 2, and $2^{1992}$ has the last digit 6.

## Answer

$9,1,2,6$.