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Divisibility Rules

  • Last Updated : 18 Mar, 2021

Suppose a boy has 537 chocolates, and he has to distribute them among his 9 friends. How can he do it? Dividing 537 by 9, and he has left with some chocolates (remainder) that mean 537 is not divisible by 9 exactly. Dividing is simple to check that the number is exactly divided by the divisor i.e. Remainder is 0 or not when one has 2 or 3-digit numbers. If the number is too large, and it takes a long time to perform the actual division. How can we know that a number is divisible by a particular divisor or not? Here comes the concept of divisibility rules: quick easy and simplest way to find out the divisibility of a number by a particular divisor.

Divisibility rule For 2

A number is divisible by 2 if the last digit of the number is any of the following digits 0, 2, 4, 6, 8.

The numbers with the last digits 0, 2, 4, 6, 8 are called even numbers, e.g. 2580, 4564, 90032 etc. are divisible by 2.

Divisibility rules for 3 and 9

A number is divisible by 3 if the sum of its digits is divisible by 3.

e.g. 90453 (9 + 0 + 4 +5 + 3 = 21) 21 is divisible by 3. 21 = 3 × 7. Therefore, 90453 is also divisible by 3. 

The same rule is also applicable to test whether the number is divisible by 9 or not but the sum of the digits of the number should divisible by 9 in the above example 90453 when we add the digits we get the result as 21 which is not divisible by 9.

e.g. 909, 5085, 8199, 9369 etc are divisible by 9. Consider 909 (9 + 0 + 9 = 18). 18 is divisible by 9(18 = 9 × 2). Therefore, 909 is also divisible by 9.

A number that is divisible by 9 also divisible by 3 but a number that is divisible by 3 not have surety that it is divisible by 9.

e.g. 18 is divisible by both 3 and 9 but 51 is divisible only by 3, can’t be divisible by 9.

Divisibility rules for 5 and 10 

A number is divisible by 5 if that last digit of that number is either 0 or 5.

e.g. 500985, 3456780, 9005643210, 12345678905 etc.

Note: A number is divisible by 10 if it has only 0 as its last digit. Eg: 89540, 3456780, 934260 etc. A number which is divisible by 10 is divisible by 5 but a number which is divisible by 5 may or may not be divisible by 10.10 is divisible by both 5 and 10 but 55 is divisible only by 5 not by 10.

Divisibility rules for 4, 6 and 8

A number is divisible by 4 if the last two digits are divisible by 4. 

e.g.: 456832960, here the last two digits are 60 that are divisible by 4 i.e.15 × 4 = 60. Therefore, the total number is divisible by 4.

A number is divisible by 6 if it is divisible by both 2 and 3.

e.g.: 10008, have 8 at one’s place so is divisible by 2 and the sum of 1, 0, 0, 0 and 8 gives the total 9 which is divisible by 3. Therefore, 10008 is divisible by 6.

  • Considering the same example let us check the divisibility rule for 8. If a number is divisible by 8 its last three digits should be divisible by 8 i.e. 008 which is divisible by 8, therefore, the total number is divisible by 8.

Divisibility rule for 11 and 7 

Consider a number to test the divisibility with 4 and 8

456832960 mark the even place values and odd place values. Sum up the digits in even place values together and sum up the digits in odd place values together.

                                                                                           

DigitsPlace Value
40
51
62
83
34
25
96
67
08

Now sum up the digits in even place values ie 0 + 2 + 4 + 6 + 8 = 4 + 6 + 3 + 9 + 0 = 22

To add up the digits in odd place values i.e.

1+ 3 + 5 + 7 = 5 + 8 + 2 + 6 = 21

Now calculate the difference between the sum of digits in even place values and the sum of digits in odd place values if the difference is divisible by 11 the complete number ie 456832960 is divisible by 11

Here the difference is 1, (22-21) it is divisible by 11. Therefore, 456832960 is divisible by 11.

Consider the number 5497555 to test if it is divisible by 7 or not.

Add the last two digits to the twice of the remaining number repeat the same process until it reduces to a two-digit number if the result obtained is divisible by 7 the number is divisible by 7.

55 + 2(54975) = 109950 + 55 = 110005

05 + 2(1100) = 2200 + 05 = 2205

05 + 2(22) = 44 + 5 = 49

Reduced to two-digit number 49 which is divisible by 7 i.e 49 = 7 × 7

Some More Divisibility Rules

Co-primes are the pair of numbers that have 1 as the common factor. If the number is divisible by such co-primes the number is also a divisible by-product of the co-primes. Eg: 80 is divisible by both 4 and 5 they are co-primes that have only 1 as the common factor, so the number is also divisible by 20 the product of 4 and 5

21 = 3 × 7                          

12 = 3 × 4

22 =11 × 2                        

14 = 2 × 7

15 = 3 × 5                            

30 = 3 × 10

18 = 2 × 9

28 = 4 × 7

26 =13 × 2.

If a number is divisible by some numbers say X that number is also divisible by factors of x. 

e.g.: if a number is divisible by 40 then it is divisible by its factors Ie: 5, 10, 2, 4, 8, 20.

Divisibility Rule For 13

If a number to be divisible by 13 add 4 times the last digit of the number to the rest of the number repeat this process until the number becomes two digits if the result is divisible by 13 then the original number is divisible by 13. 

e.g.: 333957

(4 × 7) + 33395 = 33423

(4 × 3) + 3342 = 3354

(4 × 4) + 335 = 351

(1 × 4) + 35 = 39

(1 × 4) + 35 = 39

Reduced to two-digit number 39 is divisible by 13. Therefore, 33957 is divisible by 13.

Sample Problems 

Problem 1: Determine the number divisible by 718531.

Solution:

Since, the given number contains 1 in the one’s-place, therefore it is clear that it must be divisible either by 3, 7, 9 or 11.

First add all the digits of the given number, 7+1+8+5+3+1=25 which is not divisible by 3 or 9, so 718531 is also not divisible by 3 or 9.

Lets sum up all the even places digits, 3+8+7=18

and now sum up all odd places digits, 1+5+1=7

Now subtract them as:

18-7=11

Therefore, the given number 718531 is divisible by 11.

Problem 2: Use divisibility rules to check whether 572 is divisible by 4 and 8.

Solution:

Divisibility by 4 – The last two digits of 572 is 72 (i.e. 4 x 18) is divisible by 4.

Therefore, the given number 572 is divisible by 4.

Divisibility by 8 – The last three digits of 572 is,

572 = 2 × 2 × 11 × 13

This implies that, the given number does not contain 8 as its factor, so 572 is not divisible by 8.

Problem 3: Check whether the number 21084 is divisible by 8 or not. If not then find what is that number?

Solution:

The last three digits of the given number 21084 is,

084 or 84 = 2 × 2 × 3 × 7

This implies that, the given number does not contain 8 as its factor, so 21084 is not divisible by 8.

Since, the one’s place digit of 21084 is 4 therefore it is clear that 21084 is divisible by 2.

Now, to check the divisibility by 4, consider its last two-digits: 84 i.e. 4 × 21.

This implies that, 21084 is divisible by 4.

Hence, 21084 is divisible by 2 and 4.

Problem 4: Test 224 for the divisibility by 7.

Solution:

First double the last number i.e 4 of the given number (224) ⇒ 2 × 4 = 8.

Subtract this number from the rest of the digits ⇒ 22 – 8 = 14.

This implies that, the obtained number is divisible by 7, hence the given number 224 is divisible by 7.

Problem 5: Check on 2795 for divisibility by 13.

Solution:

The last number of the given number i.e. 2795 is 5, 

Multiply 4 by 5 and add to the rest of the digits as:

⇒ 279 + (5 × 4) 

= 299.

Similarly, again multiply 4 by the last digit (i.e. 9) of the obtained three-digit number (i.e. 299) and add to the rest of the digits as:

⇒ 29 + (9 × 4) 

= 65.

Now, a two-digit number is obtained i.e. 65 = 5 × 13.

Hence, 2795 is divisible by 13.


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