Formula Sheet On Divisibility Rule
Divisibility Theorem:
Divisibility rules:
1. Divisibility by 1: All the integers are divisible by 1.
2. Divisibility by 2: A number is said to be divisible by 2 when the last digit of the
given number is even i.e. 0, 2, 4, 6, 8.
Example 1: 68, 484, 89232, 5820, 1446 all numbers are divisible by 2.
3. Divisibility by 3: A number is divisible by 3 when the sum of all the digits of the given number is either 3 or a multiple of 3.
Example 2: Check whether 78342 is divisible by 3 or not.
Solution: Sum of the digits of 78342 = 7 + 8 + 3 + 4 + 2 = 24. Here, 24 is clearly a multiple of 3. Thus, the given number 78342 will be exactly divisible by 3.
Example 3: Check whether 27353 is divisible by 3 or not.
Solution: Sum of the digits of 27353 = 2 + 7 + 3 + 5 + 3 = 20. Here, 20 is clearly not a multiple of 3. Thus, the given number 27353 will not be divisible by 3.
4. Divisibility by 4: A number is divisible by 4 when the last two digits of the given number is divisible by 4 or any multiple of 4.
Example 4: Check whether 482 is divisible by 4 or not.
Solution: Consider last two digits of 482, which is 82. Here, 82 is not divisible by 4 or multiple of 4.
Thus, 482 is not divisible by 4.
Example 5: Check whether 378244 is divisible by 4 or not. Solution:
Consider last two digits of 378244, which is 44. Here, 44 is divisible by 4 or in other words 44 is a multiple of 4. Thus, 378244 is divisible by 4.
5. Divisibility by 5: A number is divisible by 5 when the last digit of the given
the number is either 0 or 5.
Example 6: Check whether 5795 is divisible by 5 or not.
Solution: Here, the last digit of the given number is 5. Thus, 5795 is divisible by 5.
Example 7: Check whether 8732 is divisible by 5 or not.
Solution:
Here, the last digit of the given number is 2. Thus, 8732 is not divisible by 5.
We can also verify this by dividing 8732 by 5 which will leave 2 as remainder. Thus, 8732 is not divisible by 5.
6. Divisibility by 6: A number is divisible by 6 when the given number is both 2
And 3 as 6 = 2 x 3.
Example 8: Check whether 27498 is divisible by 6 or not.
Solution: Here, 27498 is exactly divisible by 2 as the last digit of the given number is even.
Also, Sum of the digits of 27498 = 2+7+4+9+8 = 30, which is divisible by 3. Thus, 27498 is also divisible by 3.
Since 27498 is divisible by both 2 and 3, the given number 27498 will be divisible by 6.
7. Divisibility by 7: There are various methods to find divisibility by 7 for any
given number. These are as following:
- Step 1: First form pairs of three-three digits from the right end of the given number.
- Step 2: Now add all the alternating pairs at odd places and even places simultaneously and find the difference between them. If the number obtained is exactly divided by 7 then the given number is said to be divisible by 7.
Example 9: Check whether 57498 is divisible by 7 or not.
Solution:
- Step 1: Form pairs of three-three digits from the right end.
Thus, 57498 is written as 057 498 (Add 0’s in the beginning of the number if necessary)
- Step 2: Since only two such are available here. Simply go for the difference between them. Difference = 498 – 057 = 441. Now check if 441 is divided by 7 or not.
Here, 441 is exactly divisible by 7. So, the given number 57498 will also be divisible by 7.
Example 10: Check whether 92384623 is divisible by 7 or not.
Solution:
- Step 1: Form pairs of three-three digits from the right end.
Thus, 92384623 is written as 092 384 623 (Add 0’s in the beginning of the number if necessary)
- Step 2: Now add all the alternating pairs at odd places and even places simultaneously and find the difference between them.
Sum of pairs at odd places = 092 + 623 = 715 Sum of pairs at even places = 384
Now, Difference = 715 – 384 = 331. Now check if 331 is divided by 7 or not.
Here, 331 is not exactly divisible by 7. So, the given number 92384623 will not be divisible by 7.
Subtract 2 times of digit at the right end from the rest of the number and repeat the process. Then check if the obtained number is divisible by 7 or not. If yes then the given number will be divisible by 7 otherwise not
Example 11: Check whether 57498 is divisible by 7 or not.
Solution:
2 times the last digit from the right end = 2 x 8 = 16
Subtract it from the rest of the number which is 5749.
Thus, 5749 – 16 = 5733
Again, 574 – (2x9) = 556
Again, 55 – (2x6)= 43
Now check 43 is divisible by 7 or not.
Since, 43 is not exactly divisible by 7 then the given number 57498 will not be divisible by 7.
8. Divisibility by 8: A number is divisible by 8 when the last three digits of the
given number is divisible by 8 or any multiple of 8.
Example 12: Check whether 274432 is divisible by 8 or not. Solution:
Here, we consider last three digits of the given number.
So, when 432 is divided by 8, it gives zero as remainder which means that 432 is completely divisible by 8. Hence 274432 will also be completely divisible by 8.
9. Divisibility by 9: A number is divisible by 9 when the sum of all the digits of
the given number is divisible by 9 or a multiple of 9.
Example 13: Check whether 873477 is divisible by 9 or not.
Solution:
Sum of the digits = 8+7+3+4+7+7 = 36
Since 36 is a multiple of 9 or is completely divisible by 9 then the given number 873477 will be exactly divisible by 9.
10. Divisibility by 10: Since 10 can be broken down into 2 multiplied by 5. So, any number that is divisible by 2 and 5 simultaneously will also be divisible by 10. Or if the last digit is 0 then the given number will be exactly divisible by 10.
Example 14:
18720 is divisible by 10 (As it is divisible by both 2 and 5).
However, 39235 is not divisible by 10 because it is completely divisible by 5 but not by 2.
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