Tests for Divisibility of Numbers
Is the number 38 divisible by 2? by 4? by 5?
By actually dividing 38 by these numbers we find that it is divisible by 2 but not by 4 and by 5.
Let us see whether we can find a pattern that can tell us whether a number is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10 or 11. Do you think such patterns can be easily seen?
Divisibility by 10
Charu was looking at the multiples of 10.
The multiples are 10, 20, 30, 40, 50, 60, ... . She found something common in these numbers. Can you tell what? Each of these numbers has 0 in the ones place.
She thought of some more numbers with 0 at ones place like 100, 1000, 3200, 7010.
She also found that all such numbers are divisible by 10.
She finds that if a number has 0 in the ones place then it is divisible by 10.
Can you find out the divisibility rule for 100?
There are a few different rules that can make it surprisingly easy to check if a number is divisible by another. In this section we will have a look at some of them…
Divisibility by 2, 5 and 10
Every number is divisible by 1. To determine if a number is divisible by 2, we simply have to check if it’s even: any number that ends in 0, 2, 4, 6, or 8 is divisible by 2.
Charu observes a few multiples of 2 to be 10, 12, 14, 16... and also numbers like 2410, 4356, 1358, 2972, 5974.
She finds some pattern in the ones place of these numbers.
Can you tell what?
These numbers have only the digits 0, 2, 4, 6, 8 in the ones place.
She divides these numbers by 2 and gets remainder 0.
She also finds that the numbers 2467, 4829 are not divisible by 2. These numbers do not have 0, 2, 4, 6 or 8 in their ones place.
Looking at these observations she concludes that a number is divisible by 2 if it has any of the digits 0, 2, 4, 6 or 8 in its ones place.
Mani found some interesting pattern in the numbers 5, 10, 15, 20, 25, 30, 35, ...
Can you tell the pattern? Look at the units place.
All these numbers have either 0 or 5 in their ones place. We know that these numbers are divisible by 5.
Mani took up some more numbers that are divisible by 5, like 105, 215, 6205, 3500.
Again these numbers have either 0 or 5 in their ones places.
He tried to divide the numbers 23, 56, 97 by 5. Will he be able to do that? Check it.
He observes that a number which has either 0 or 5 in its ones place is divisible by 5, other numbers leave a remainder.
Is 1750125 divisible 5?
To see if a number is divisible by 5, we similarly just have to check that its last digit is 0 or 5:
The reason why these rules for 2 and 5 are so simple has to do with our number system. The base of our number system is 10, which means that every digit in a number is worth 10 times as much as the next one to the right. If we take the number 6382 as an example,
| 6 | 3 | 8 | 2 |
| =6000 | =300 | =80 | =2 |
Now we can separate the last digit of a number from all its other digits:
| abcd | = | abc × 10 | + | d |
| 6382 | = | 638 × 10 | + | 2 |
Both 2 and 5 are factors of 10, so they will
The easiest is the divisibility rule for 10: we just need to check if the
Divisibility by 4 and 8
Unfortunately 4 doesn’t divide 10, so we can’t just look at the last number – but 4 does divide 100, so we just have to slightly modify our rule from above. Now we write ab cd = ab × 100 + cd. We know that 4 will always divide ab × 100, so we have to look at the last
For example, 24 is divisible by 4 so 2735 24
Can you quickly give five 3-digit numbers divisible by 4? One such number is 212. Think of such 4-digit numbers.
One example is 1936.
Observe the number formed by the ones and tens places of 212. It is 12;
which is divisible by 4. For 1936 it is 36, again divisible by 4.
Try the exercise with other such numbers, for example with 4612; 3516; 9532.
Is the number 286 divisible by 4?
So, we see that a number with 3 or more digits is divisible by 4 if the number formed by its last two digits (i.e. ones and tens) is divisible by 4.
Check this rule by taking ten more examples.
Divisibility for 1 or 2 digit numbers by 4 has to be checked by actual division.
The divisibility rules for 8 gets even more difficult, because 100 is not divisible by 8. Instead we have to go up to
For example, 120 is divisible by 8 so 271120 is
Divisibility by 3 and 9
The divisibility rule for 3 is rather more difficult. 3 doesn’t divide 10, and it also doesn’t divide 100, or 1000, or any larger power of 10. Simply looking at the last few digits of a number isn’t going to work.
Instead we need to use the digit sum of a number, which is simply the sum of all its individual digits. For example, the digit sum of
Are the numbers 21, 27, 36, 54, 219 divisible by 3?
Are the numbers 25, 37, 260 divisible by 3?
Can you see any pattern in the ones place? We cannot, because numbers with the same digit in the ones places can be divisible by 3, like 27, or may not be divisible by 3 like 17, 37. Let us now try to add the digits of 21, 36, 54 and 219. Do you observe anything special ? 2+1=
All these additions are divisible by 3.
Add the digits in 25, 37, 260. We get 2+5=
These are not divisible by 3.
We say that if the sum of the digits is a multiple of 3, then the number is divisible by 3.
Is 7221 divisible by 3?
Here we’ve highlighted all numbers which are multiples of three. You can see that their digit sums are always
So to determine if any number is divisible by 3, you just have to calculate its digit sum, and check if the result is also divisible by 3.
Next, let’s look at multiples of 9:
It seems that all the numbers divisible by 9 have a digit sum which is
The multiples of 9 are 9, 18, 27, 36, 45, 54,...
There are other numbers like 4608, 5283 that are also divisible by 9.
Do you find any pattern when the digits of these numbers are added?
1 + 8 =
4 + 6 + 0 + 8 =
All these sums are also divisible by 9.
Is the number 758 divisible by 9?
The sum of its digits 7 + 5 + 8 = 20 is also not divisible by 9.
These observations lead us to say that if the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.
Of course, these curious patterns for numbers divisible by 3 and 9 must have some reason – and like before it has to do with our base 10 numbers system. As we saw, writing the number 6 3 8 4 really means
6 × 1000 + 3 × 100 + 8 × 10 + 4.
We can split up each of these products into two parts:
6 × 999 + 6 + 3 × 99 + 3 + 8 × 9 + 8 + 4.
Of course, 9, 99, 999, and so on are always divisible by 3 (or by 9). All that remains is to check that what’s left over is also divisible by 3 (or 9):
6 + 3 + 8 + 4
This just happens to be the digit sum! So if the
Divisibility by 6
We’ve still skipped number 6 – but we’ve already done all the hard work. Remember that 6 = 2 × 3.
To check if a number is divisible by 6 we just have to check that it is divisible by 2
Can you identify a number which is divisible by both 2 and 3? One such number is 18. Will 18 be divisible by 2×3=
Find some more numbers like 18 and check if they are divisible by 6 also.
Can you quickly think of a number which is divisible by 2 but not by 3?
Now for a number divisible by 3 but not by 2, one example is 27.
Is 27 divisible by 6?
From these observations we conclude that if a number is divisible by 2 and 3 both then it is divisible by 6 also.
Divisibility by 7
Let's take the number 469. Take the last digit, which is 9. Double it: 9 * 2 =
Subtract 18 from the rest of the number without the last digit (46): 46 - 18 =
Now, check if 28 is divisible by 7. It is, because 28 divided by 7 equals 4.
Therefore, we can say that 469 is divisible by
Consider the following numbers which are divisible by 7: 133, 273, 329, 595, 672. Check these numbers are divisible by 7 using the divisibility rule of 7.
In the number 133, double the last digit of the number 3, which gives 6. Now we need to subtract it from the rest of the remaining number, i.e. 13 – 6 =
Since 7 is divisible by 7. Hence, 133 is also divisible by 7.
In the number 273, double the last digit of the number 3, which gives 6. Now we need to subtract it from the rest of the remaining number, i.e.
Since 21 is divisible by 7. Hence, 273 is also divisible by 7.
In the number 329, double the last digit of the number 9, which gives 18. Now we need to subtract it from the rest of the remaining number, i.e. 32 –
Since 14 is divisible by 7. Hence, 329 is also divisible by 7.
In the number 595, double the last digit of the number 5, which gives 10. Now we need to subtract it from the rest of the remaining number, i.e. 59 – 10 =
Since 49 is divisible by 7. Hence,
In the number 672, double the last digit of the number 2, which gives 4. Now we need to subtract it from the rest of the remaining number, i.e. 67 – 4 =
Since 63 is divisible by 7. Hence, 672 is also divisible by
Divisibility by 11
The numbers 308, 1331 and 61809 are all divisible by 11. We form a table and see if the digits in these numbers lead us to some pattern.
| Number | Sum of the digits (at odd places) from the right | Sum of the digits (at even places) from the right | Difference |
|---|---|---|---|
| 308 | 8 + 3 = | 0 | 11 – 0 = |
| 1331 | 1 + 3 = | 3 + 1 = 4 | 4 – 4 = |
| 61809 | 9 + 8 + 6 = | 0 + 1 = 1 | 23 – 1 = |
We observe that in each case the difference is either 0 or divisible by 11. All these numbers are also divisible by 11.
For the number 5081, the difference of the digits is (5+8) – (1+0) =
which is not divisible by 11. The number 5081 is also not divisible by 11.
Thus, to check the divisibility of a number by 11, the rule is,
find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number. If the difference is either 0 or divisible by 11, then the number is divisible by 11.