This is post is the fourth part of the Divisibility Rules Series. The first three posts are about divisibility by 2, divisibility by 4 and divisibility by 5 and 10. In this post, we discuss divisibility by 3.

When we divide things, we split them into ‘groups.’ Of course, we cannot always divide them equally. For example, dividing 13 balls by 3 is the same as putting the balls into containers in groups of three. In dividing 13 by 3, 1 is left out as shown below. We have learned that the excess or the ones that are left out during division are called *remainders*. In the figure below, we can say that 13 divided by 3 gives a remainder of 1.

Divisibility by 3 is not as easy as divisibility by 2, by 5 or by 10. However, we note the following observations when we divide by 3.

First, we can observe that all powers of 10 when divided by three give a remainder of one. Ten divided by 3 gives a remainder of 1, 100 divided by three gives a remainder of 1, 1000 divided by 3 gives a remainder of 1 and so on. Can you think of a reason why is this so?

Second, we can see that any number multiplied by any power of 10 gives the same remainder when divided by 3. The numbers 7, 70, 700, 7000 and so on gives a remainder of 1 when divided by 3. From this observation, we can conclude that the place value of a number does not matter with regard to divisibility by 3. We can look at the numbers as digits and not considering their place value. For example, 366 is divisible by 3 because all the digits are divisible by 3. We can easily verify that because 300, 60, and 6 are all divisible by 3. Now, how about 2673? Is it divisible by 3? Only two digits are divisible by 3; those are 6 and 3*. *

Since we are looking at the digits and not considering the place value, we can evaluate each digit and examine the remainders. As mentioned above, 6 and 3 are divisibly by 3, because 600 and 3 are divisible by 3. Two however (or 2000) gives 2 as place value while 7 (or 70) gives 1 as remainder when divided by 3. Notice that if we use the illustration above, then we can combine the remaining balls (remainders) from the two operations and put them in a container. In effect, there is no excess. Therefore, the 2673 is divisible by 3.

Notice that dividing the digits individually and combining them first before dividing would result to the same thing. If we are searching for 3-ball containers that would contain 2 balls (from the thousands digit) and the 7 balls (from the tens digits) is the same searching for containers for nine balls (7+2). Therefore, if we combine all the balls and put them into 3-ball containers, we would see if there are remainders. This means that we can add all the digits of a number and if the sum is divisible by 3, then it is divisible by 3. Hence, we have the following conclustion.

**A number is divisible by 3, if the sum of their digits is divisible by 3.**

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Hi,

Nice piece! I was considering linking to it. However, for my readers’ sakes, I think you should explain the following in more detail, and not rush through it as an easy “observation”:

“Second, we can see that any number multiplied by any power of 10 gives the same remainder when divided by 3. The numbers 7, 70, 700, 7000 and so on gives a remainder of 1 when divided by 3.”

Thanks Maria. Yes, I will try to explain further in a while.