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Divisibility Rules (For Decimal System)

Divisibility rules are quite imperative, because with the help of these rules we can infer if a particular number is divisible by another number or not, without actually dividing it. Divisibility rules of numbers are specific to that particular number only. It simply means that divisibility rules of different numbers will be different. We shall now see a list of divisibility rules for some of the natural numbers.

Divisibility Rules

For 2 If unit digit of any number is 0, 2, 4, 6 or 8, then that number will be divisible by 2.
For 3 If sum total of all the digits of any number is divisible by 3, then the number will be divisible by 3. (e.g., 123, 456 etc.)
 
Example-1
How many values of A are possible if 3245684A is divisible by 3?
Solution
Sum total of the number = 32 + A
For this number to be divisible by 3, A can take three values, namely, 1 or 4 or 7. (No other values are possible since A is the unit digit of the number.)
 

For 4 If last two digits of a number are divisible by 4, then that number will be divisible by 4 (e.g., 3796, 248, 1256).
 
For 5 If last digit of the number is 5 or 0, then that number will be divisible by 5.
 
For 6 If last digit of the number is divisible by two and sum total of all the digits of the number is divisible by 3, then that number will be divisible by 6.
 
For 7 The integers are divisible by 7 if and only if, the difference of the number of its thousands and the remainder of its divisible by 1000, is divisible by 7.
 
For 7 If the difference between the numbers of tens in the number and twice the unit digits is divisible by 7, then the given number is divisible by 7.
E.g., Let us take the number 795. The units digit is 5 and when it is doubled it, we get 10. The remaining part of the number (i.e., the tens) is 79. If 10 is subtracted from 79 we get 69. Since this result is not divisible by 7, the original number 795 is not divisible by 7.
 
For 8 If last 3 digits of the number are divisible by 8, then the number itself will be divisible by 8 e.g., 128, 34568, 76232 etc.
 
For 9 If sum of the digits of the number is divisible by 9, then the number will be divisible by 9 e.g., 129835782.
 
1 + 2 + 9 + 8 + 3 + 5 + 7 + 8 + 2 = 45. Since 45 is divisible by 9, then the number will be divisible by 9.
 
Example-2
How many pairs of A and B are possible in number 89765A4B if the number is divisible by 9, given that last digit of the number is even?
Solution
Sum of digits of number is 8 + 9 + 7 + 6 + 5 + A + 4 + B = 39 + A + B.
So, A + B should be 6 or 15. Next value should be 24 but since A and B are digits so it can’t be more than 18. Possible pairs of A and B are
 
A B
0 6
1 5
2 4
3 3
4 2
5 1
6 0
7 8
8 7
9 6
6 9

Since B is even, six possible set of values of A and B are there.

 

For 11 A number is divisible by 11, if the difference between the sum of the digits at the even places and the sum of the digits at the odd places is divisible by 11 (zero is divisible by 11) e.g., 6595149 is divisible by 11 as the difference of 6 + 9 + 1 + 9 = 25 and 5 + 5 + 4 = 14 is 11.
 
For 12 If the number is divisible by 3 and 4, then the number will be divisible by 12, e.g., 144, 348.
 
For 13 (A + 4B), where B is the units place digit and A is all the remaining digits.
E.g., Checking the divisibility of 1404 by 13: Here, A =140 and B = 4, then A+4B=140 +4×4=156. 156 is divisible by 13, so 1404 will be divisible by 13.
 
For 14 If the number is divisible by both 2 and 7, then the number will be divisible by 14.
 
For 15 A number is divisible by 15, if the sum of the digits is divisible by 3 and unit digit of the number is 0 or 5, e.g., 225, 450, 375 etc.
 
For 16 A number is divisible by 16, if the number formed by the last 4 digits of the given number is divisible by 16 e.g., 12578320 is divisible by 16, since last 4 digits of the number, 8320 is divisible by 16.
 
For 17 (A – 5B), where B is the units place digit and A is all the remaining digits.
 
For 18 Number should be divisible by both 9 and 2.
For 19 (A + 2B), where B is the units place digit and A is all the remaining digits.
 
If the sum of the number of tens in the number and twice the unit digit is divisible by 19, then the number is divisible by 19.
 
For example, let us take the number 665. The units digit is 5 and when it is doubled, we get 10. The remaining part of the number is 66. If 10 (which is the unit digit doubled) is added to 66 we get 76. Since, 76 is divisible by 19, it means the original number 665 is also divisible by 19.
 
For 20 The number should be divisible by 4 and 5.

Process to find out the divisibility rule for prime numbers

Process for the divisibility rule for P, a prime number is as follows:
 
Step 1 Find the multiple of P, closest to any multiple of 10. (This essentially will be of the form 10K + 1 or 10K –1.)
 
Step 2 If it is 10 K – 1, then the divisibility rule will be A + KB, and if it is 10K + 1, then the divisibility rule will be A – KB, where B is the units place digit and A is all the remaining digits.
E.g., Finding out the divisibility rule of 23: Lowest Multiple of 23, which is closest to any multiple of
10 = 69 = 7 × 10 – 1
So, rule is A + 7 B.

Divisibility rule for any number of the format

10n ± 1, where n is any natural number

In this method, we will be considering two factors, the first one being the value of n and the second one being the sign ‘+’ or ‘’. The value of ‘n’ signifies how many digits will be taken one at a time and the sign signifies in what manner these digits will be taken. Let us see this with the help of some examples.
 

When n = 1
 
 
Case-1 10n + 1 = 11
This method tells us that now we will be considering one digit at a time of the number which is to be divided by 11 and since the sign is ‘+’ between 10n and 1, we will alternately subtract and add starting from the right hand side.
 
E.g., To check if 123345 is divisible by 11, we will start from the right hand side, by subtracting the first number and then adding the third and then subtracting the fourth and so on…

Case-2 10n – 1 = 9
Since the value of n is one, we will take one digit at a time and due to the ‘–’ sign between 10n and 1, we will keep on adding the digits from the right hand side (or, can be done from the left hand side also in this case).

When n = 2
 
Case-3 10n + 1 = 101
Since the value of n here is two, we will make the pairs of digits from right hand side, and then we will alternately subtract and add each of the pairs.
 
E.g., Checking if 12345678 is divisible by 101: Make the pairs first – 12 34 56 78. Now we will do this calculation – (78) – (56) + (34) – (12) = 44, which is not a multiple of 101, so this number 12345678 is not divisible by 101.

Case-4 10n  1 = 99
Since the value of n is two, we will take two digit at a time after making the pairs and due to the ‘–’ sign between 10n and 1, we will keep on adding the pairs from right hand side (or, can be done from left hand side also in this case).
 
E.g., Checking if 122166 is divisible by 99: Let us make the pairs first: 12 21 66. Now add the pairs 12 + 21 + 66 = 99, so this number 122166 is divisible by 99.
 
And so on, we extend it for any natural number value of n.




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