The origins of these tests can be dated back to before the Renaissance and well before the invention of decimals because of their usefulness in reducing large number fractions to their lowest terms. At this level we won't be too concerned with how these rules work, suffice it to say we will explain them by examples.


Divisibility by 2


We know that if a number is even it is divisible by 2, so any even number will automatically be divisible (at least once) by 2. We don't need to do any examples of this because hopefully it is quite obvious.


Divisibility by 3


For a number to be divisible by 3, we add up all of the individual digits making up the number, and this will bring us to another one. If this number that we end up with contains more than one digit, we once again and these up and keep repeating the process until we end up with a number which contains only a single digit. If that number is divisible by 3, then the original number was also divisible by 3. The proof of this can be found in many advanced level mathematics "number theory" texts but, like I said we will just be satisfied with the fact that it works and take a look at the divisibility of a few numbers just to confirm this.


Q. Is the number 643 divisible by 3?


A. Add up the numbers 6, 4 and 3 to arrive at 13. Add up the numbers 1 and 3 to arrive at 4. Since the number 4 itself is not divisible by 3, then neither was the original number 643.


Q. Is the number 2487 divisible by 3?


A. Add up the individual digits 2, 4, 8 and 7 to arrive at 21, add 2 and 1 to make 3. The number 3 is obviously divisible by 3, therefore 2487 was also divisible by 3


Q. Is the number 64,238,973,842 divisible by 3?


A. Add up the digits in just the same way and try not to be put off by such a massive number. The total of 6, 4, 2, 3, 8, 9, 7, 3, 8, 4 and 2 = 56. 5+6 is 11 and this itself is not divisible by 3, therefore the original number 64,238,973,842 was not divisible by 3


Divisibility by 4


For a number to be divisible by 4, we take the last 2 digits and see if they on their own make a number which is divisible by 4, if they do then the original number was also divisible by 4


Q. Is 298 divisible by 4?


A. The last 2 numbers make up the number 98, and although this is even and divisible by 2 it cannot be divided integrally by 4 (in other words you can't divide by 4 without coming back with an answer which is not a whole number). therefore 298 is not divisible by 4.


Q. Is 29,180 divisible by 4?


A. You can see that the last 2 digits form the number 80 which hopefully you will know is clearly divisible by 4 (20), therefore the original number 29,180 is also divisible by 4.


Q. Is 24,284,936 divisible by 4?


A. By now you should be able to see that as the last 2 digits form the number 36 which is divisible by 4, then the original number 24,284,936 is also divisible by 4.


Divisibility by 5


For a number to be divisible by 5 it has to end with either a 5 or a 0, this is one of the easier tests to apply.


Divisibility by 6


For a number to be divisible by 6 it has to be divisible by 2 and also by 3. The first test is usually quite easy to apply, if the number is even then it will be divisible by 2, and if the sum of the individual digits comes to a number which is divisible by 3 than the number itself is also divisible by 3. If we combine both of these rules then the original number would itself be divisible by 6.


Q. Is the number 26,404 divisible by 6?


A. You should be able to see that the number is divisible by 2 by virtue of the fact that it is even, but is it also divisible by 3? Let us add up the digits and see:


2+6+4+0+4=16, 1+6= 7


This tells us that the number is not divisible by 6 because the sum of the individual digits does not divide by 3, remember to be divisible by 6 the number has to conform to both divisibility rules.


Q. Is the number 26,406 divisible by 6?


A. You can see that the number is divisible by 2 because it is even, let us add up the individual digits and see that results in the number that is divisible by 3:


2+6+4+0+6 = 18 and 1 + 8 = 9


Ultimately the digits sum is divisible by 3 (9 does indeed divide by 3) therefore as both rules are satisfied, the original number 26,406 is divisible by 6.


Q. Is the number 28,648,248,222 divisible by 6?


A. Straight away you can see that the first part of the divisibility rule would be satisfied because this is an even number and would therefore be divisible by 2. But what is the digit sum?


2+8+6+4+8+2+4+8+2+2+2 = 48 and 4 + 8 = 12 and 1 + 2 = 3


The final digit sum is 3 which is a number divisible by 3 :-) therefore we know that the original large number would also be divisible by 6.


Divisibility by 7


The rule that we need to apply to explore divisibility by 7 is quite unique. Although the arithmetic involved isn't particularly complicated, it can be quite fiddly so you will really need to concentrate on this. So, how do we tell if a number is divisible by 7?


151263


Step 1 - look at the number and physically remove the last digit:


15126


Step 2 - double the number that you took away and subtract its result from the number that you have left. So in this case we took away 3, and if we double this it makes 6, then we take this from 15126 which leaves us with 15120.


15120


Now repeat step 2 until it can no longer be repeated. Removing the 0 from 15120 leaves us with 1512


1512


If we multiply the number that we took away (0) by 2 and remove it from 1512 we remain with 1512. Now we repeat the step again by taking away 2 leaving 151:


151


Double the number that we took away (2×2 = 4) and subtract this from 151 leaving us 147:


147


Repeat the step again, taking 7 away from 147 leaving us with 14:


14


This final result gives us a number which is clearly/visibly divisible by 7 (14÷7 = 2) so we conclude that original number was also divisible by 7.


Q. Is the number 75,269 divisible by 7?


A. Just follow the rules as we did previously and remove the 9 from the number given leaving us with 7526. Double the 6 making 12 and remove this from 7526 leaving us with 7514:


7514


Repeat the step by taking away the 4 to leave us with 751, double it making 8 and subtract it from 751 leaving us with 743:


743


Remove the 3, double it and subtract it from 74 leaving us with 68.


68


Remove the 8, double it making 16 and subtract it from 68 which leaves us with 52.


We can stop at this point, as the number 52 is not recognisable as a number which obviously divides by 7 (the closest numbers that do in fact divide by 7 are 49 and 56) so we can safely say that our original number was not divisible by 7.


Divisibility by 8


This is a slightly simpler rule than the one previously you will be glad to know. If we take a number and look at the last 3 digits and can see a number comprised of those digits which divides by 8 then we conclude that the original number is also divisible by 8.


Q. Is the number 42987 divisible by 8?


A. The last 3 numbers are 987, a number which is not divisible by 8 therefore the original number of 42987 also would not been divisible by 8.


Q. Is the number 4042752 divisible by 8?


A. The last 3 digits form the number 752 which is divisible by 8 (752÷8 = 94) therefore the original number 4042752 was also divisible by 8.


Divisibility by 9


The divisibility rule for 9 is very similar to that for 3. If we take a number and add up its constituent digits to arrive at a number that is divisible by 9, then the original number was also divisible by 9.


Q. Is the number 42,803,964 divisible by 9?


A. Let us take the sum of the individual numbers:


4+2+8+0+3+9+6+4 = 36 and 3 + 6 = 9. As 9 is clearly divisible by 9 then so was the original number!.


Q. Is the number 38,880 divisible by 9?


A. 3+8+8+8+0 = 27 and 2+7 = 9 therefore the original number must also have been divisible by 9


Divisibility by 10


I don't think there is any need to give any examples of this, any number ending in a 0 is divisible by 10.


Divisibility by 11


This division test is probably one of the weirdest ones that you can come across. To determine whether or not a number is divisible by 11 we look at the number and total all of the alternating digits. Doing this will provide us with 2 numbers and if the difference of these 2 numbers is divisible by 11 then so was the original number. This will hopefully make a little bit more sense when you have seen a couple of examples:


Q. Is the number 12345 divisible by 11?


A. Let us add up the numbers that would be in the "odd" columns if we were to write the number out in numbered columns and then do the same thing for those numbers which would have been in the "even" columns:


1+3+5 = 9

2+4 = 6


Now we take the difference (9-6=3) and we can see that the resulting number 3 is not divisible by 11, therefore 12345 was also not divisible by 11. Now let's take a look at an example which is.


Q. Is 57,728 divisible by 11?


A. Divide the integers forming the number into those which would fall into the "odd" columns and those which would fall into the "even" columns:


5+7+8 = 20

7+2 = 9


We can see that if we subtract 9 from 20 we are in fact left with 11. Clearly 11 is divisible by 11 and so the original number 57,728 would have also been divisible by 11.


Let's go out with a bang :-)


Q. Is the number 12,422,185,732 divisible by 11?


A. Simply group the numbers accordingly and see if the resulting expression is divisible by 11, if it is then so is the number if not then neither is the number.


1+4+2+8+7+2 = 24

2+2+1+5+3 = 13


24 - 13 = 11


I think we can safely conclude that our huge number to start with was divisible by 11 :-)