The last digit of a number is the digit at the very end, also called the unit digit. In this lesson, we will focus on how to find out the last digit of a calculation, without knowing what the whole answer is.

Imagine that you are going to add two numbers. For example $744 + 418$. The most common way to do this by hand is to do long addition. You write the numbers on top of each other and then add the numbers digit by digit from the right to left. You probably notice that you know the last digit of the sum already when you have added $4$ with $8$ and gotten $12$. So you will have written that the sum ends in $2$ and carried the $1$ to the tens place. The important thing here is that we can know that $744 + 418$ ends in $2$, without doing the full calculation and knowing that the sum computes to $1162$. We can do this by looking at the sum of the last digits, $4 + 8 = 12$ and see that this ends in $2$.

Why is it like that? The example actually shows this quite clearly. When we added the last digits we got $12$, with $1$ carrying over to the next digit. We only have digits that carry from right to left, never from left to right. Therefore, only the last digits affect what the last digit in the answer will be.

Another way to illustrate this is to split off the last digit from the numbers we add. For example, $744 = 740 + 4$ and $418 = 410 + 8$. Then $744 + 418 = (740 + 4) + (410 + 8)$. Since $740$ and $410$ both end in $0$, they do not affect the last digit. So it is enough to look at the last digit of the sum $4 + 8$.

The same idea works for multiplication. If you want to multiply two large numbers $744 \cdot 418$, it is enough to look at the product of the last digits, $4 \cdot 8 = 32$, and see that the last digit becomes $2$. Feel free to think about what this would look like if you did the multiplication by hand with pen and paper.

To illustrate this, we can do in the same way as for addition. By dividing the numbers as $744 = 740 + 4$ and $418 = 410 + 8$. Now $744 \cdot 418$ becomes $(740 + 4) \cdot (410 + 8)$. This is the same as $740 \cdot 410 + 740 \cdot 8 + 4 \cdot 410 + 4 \cdot 8$. Here, all terms except $4 \cdot 8$ contain a factor ending in $0$, and will therefore not affect the last digit.

OBS! The corresponding does not work for division! For example, $\frac{12}{2} = 6$, but if we do the same calculation with the last digits, we get $\frac{2}{2} = 1$. This has to do with the fact that it doesn't make sense to split up $12$ as $10 ​​+ 2$ because $\frac{10}{2} = 5$ doesn't end in $0$. Examining only last digits generally doesn't work with division.

We will now go through how to calculate the last digit for powers such as $13^8$. If you do not know what powers are, you can briefly say that powers are repeated multiplication. $13^8 = 13 \cdot 13 \cdot 13 \dots 13$, in total a count of eight $13$ times each other.

A first simplification is that $13^8$ has the same last digit as $3^8$. We can understand this with the argument for last digits in multiplication that we learned earlier. If we multiply $13 \cdot 13 \cdot 13 \dots 13$, it will have the same last digit as if we multiply $3 \cdot 3 \cdot 3 \dots 3$.

After that, we can do the multiplication step by step and delete the numbers that are not last numbers because they will not affect the last number in the answer.

What is the last digit in $13^8$?

Maybe you noticed that the last digits in the example are repeated. The last digits were $3, 9, 7, 1, 3, 9, 7, 1$. If we think about it it is only the previous last digit that determines what the next last digit will be, since we take that number times $3$. Then we could really as soon as we saw that $3 ^ 5$ had the last digit $3$, the same as $3 ^ 1$ understand that the last digit would be repeated as $3, 9, 7, 1$ over and over again. We can use this to calculate the last digit of even larger powers, such as $13 ^ {501}$. Since it is a repeating cycle, we only need to know where in the cycle $13 ^ {501}$ ends up. The digits are repeated in a cycle of length $4$ and we wonder what number $501$ is in that cycle. We can find out with a remainder from division. If we divide $501$ with $4$ we get $\frac {501} {4} = 125$ and remainder $1$. We can see this as going $125$ laps and then one more step. Then we end up with the first element in the cycle, $3$. Another way of looking at it is that the last digit for $13 ^ {501}$ is the same as $13 ^ 1$, just $125$ rounds earlier, which has last digit $3$.

The method for finding the last digit for large powers is thus to calculate what the initial last digits will be, until they start to repeat themselves. They will always eventually repeat because there are only $10$ different last digits possible and as soon as the same last digit has appeared twice, the digits will repeat as they did in between. When you know what digits are included in the cycle and how long it lasts before it repeats itself, the last step is to find out where in the cycle the power you are looking for ends up. We do this with the help of a remainder from division. For example if a cycle is $4$ digits long and the power is $247$ after removing the part before the cycle starts (if there is such a part) we just need to divide $247$ with $4$ and see that the remainder is $3$ to know that it is the third digit in the cycle.