What are the Multiples of 8?

The number eight is regarded as a lucky number in Chinese and other Asian cultures because it sounds like generating wealth. According to Pythagorean numerology, eight also means victory, prosperity, and overcoming.

It’s incredible how numbers can be this lucky and meaningful. Now, let’s try to learn more about this number and its multiple!

Multiples of 8

A multiple of 8 is a sequence in which the difference between each next number and the previous number is 8. You can notice it given some of the multiples of 8 like 8, 16, 24, 32, 40, 48, 56, 64, 72, 80… and so on.

Furthermore, when the numbers are divided by 8, the result should always be an exact number. To put it another way, a multiple of 8 is the result of multiplying 8 and any number.

If we multiply 8 to any positive number, we will get a positive multiple of 8. Similarly, when we multiply 8 to any negative integer, we get a negative multiple of 8. It is worth noting that fractions cannot be used to generate a multiple of 8, since we need an exact number.

How to determine the multiples of 8?

Now that we have defined what it means to be a multiple of 8, let’s proceed with finding the possible multiples of 8.

One way to find the multiples of 8 is through repeated addition or skip counting. This process is done by simply adding 8 as many times as necessary.

The figure shows how we can do skip counting or repeated addition. First, we start by putting 8 on top of the “ladder” because 8 is the smallest multiple of 8.

By repeated addition, we will simply add 8 as many times as we want. Hence, $$8\;+\;8\;=\;16$$,

$$16\;+\;8\;=\;24$$

… and so on.

We have defined multiples of 8 as any number that can be expressed as 8n, where n is an integer and a multiple of 8, is a multiple of 8. Thus, a different approach to find any multiples of 8 is by multiplying 8 to any integer.

For example, to find the 15^th multiple of 8, all we need to do is multiply 8 by 15. Hence, $$8\;\times\;15\;=\;120$$. Therefore, the 15^th multiple of 8 is 120.

Let’s try to get another multiple of 8. This time, let’s look for the 112^th negative multiple of 8. Since we are looking for the negative multiple of 8, we need to multiply 8 by -112. Thus, $$8\;\times\left(-\;112\right)=-896$$.

Now, let’s look at this table.

nth Multiple	Skip Counting	Multiplication
1st multiple	8	8 x 1 = 8
2nd multiple	8 + 8 = 16	8 x 2 = 16
3rd multiple	8 + 8 + 8 = 24	8 x 3 = 24
4th multiple	8 + 8 + 8 + 8 = 32	8 x 4 = 32
5th multiple	8 + 8 + 8 + 8 + 8 = 40	8 x 5 = 40

See how the result in skip-counting and multiplication are the same? That’s because the multiplication process is an alternative and faster way to get multiples of 8.

List of the First 30 Multiples of 8

The table shows the first 30 positive multiples of 8 and its factors. We can clearly see that if we multiply any number to 8, it will also generate different multiples of 8.

Product of 8 and a positive counting number	Multiples of 8
8 x 1	8
8 x 2	16
8 x 3	24
8 x 4	32
8 x 5	40
8 x 6	48
8 x 7	56
8 x 8	64
8 x 9	72
8 x 10	80
8 x 11	88
8 x 12	96
8 x 13	104
8 x 14	112
8 x 15	120
8 x 16	128
8 x 17	136
8 x 18	144
8 x 19	152
8 x 20	160
8 x 21	168
8 x 22	176
8 x 23	184
8 x 24	192
8 x 25	200
8 x 26	208
8 x 27	216
8 x 28	224
8 x 29	232
8 x 30	240

Now, do you see any pattern on the multiples of 8? If we take a closer look at the unit’s digit of the multiples of 8, it follows the pattern 8, 6, 4, 2, and 0.

Did you know that…

We can easily determine if a large number is a multiple of 8? This way, we do not need to divide a large number by 8. This can be done by simply getting the last three digits of a number and dividing it by 8. If the last three digits of any number are divisible by 8, we are certain that it is a multiple of 8.

Amazing, right? Now let’s see if this discovery is true.

Say, for example, the number 1864. The last three digits of 1864 are 864. If we divide 864 by 8, we will get a result of 108 – which means 864 is a multiple of 8. Now, if we divide 1864 by 8, we will get an exact result of 233.

We are now sure that if the last three-digit number of a large number is a multiple of 8, we can already conclude that the number is also a multiple of 8.

Can you try it with 23 112 and 5 649 480?

Problems involving multiples of 8

Now, let’s apply what we have learned about the multiples of 8 by solving these real-life problems.

Problem #1

Amirah promised her friends that she would pay for their shopping if the price of the item is a multiple of 8. Joan got a pair of jeans for $144, and Sheryl got a top for $56. Will Amirah pay for their clothes?

To know if Amirah will pay for Joan and Sheryl’s items, we need to determine if $144 and $56 are multiples of 8.

Let us first determine if Amirah will pay for the pair of jeans Joan got. Since the price of the jeans is $144, we need to divide it by 8 to check if it is a multiple of 8. Hence,

$$144\;\div\;8\;=\;18$$.

Since we got a zero remainder, we know that Amirah will pay for Joan’s jeans.

Now, let’s check if she will also pay for Sheryl’s top. The price of the top that Sheryl got is $56. If we divide 56 by 8, we will get the result

$$56\;\div\;8\;=\;7$$.

Since 56 is a multiple of 8, we can conclude that Amirah will also pay for Sheryl’s item.

Therefore, Amirah will pay for the items Joan and Sheryl picked.

Now, let’s try another problem.

Problem #2

Emmy is trying to learn English by writing new English words in her notebook. She wrote 8 English words on her first day, 16 on the second, 24 on the third… and so on. How many new English words will she write by the end of the year?

To answer this problem, we need to know the number of days in a year. There are 365 days in a year and 366 days if it is a leap year. If Emmy started to learn English in a non-leap year, we need to multiply 8 by 365. Hence,

$$8\;\times\;365\;=\;2920$$.

Thus, Emmy needs to write 2920 new English words by the end of a non-leap year.

If Emmy started studying English on a leap year, we need to multiply 8 by 366. Thus,

$$8\;\times\;366\;=\;2928$$.

Hence, she needs to write 2928 new English words by the end of a leap year.

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