HOW CAN V CHECK A NUMBER WHETHER IT IS DIVISIBLE BY 2,3,4,5,6,8,9,10,11?
In order to find whether a number is divisible by 2, 3,4,5,6,8,9,10,11, we have to check the divisibility test for the number. Divisibility tests are given below:
1. A number is divisible by 2, if it has any of the digits 0, 2, 4, 6 or 8 at its ones place, i.e., it is an even number. (Examples: 10, 12, 14, .....)
2. A number is divisible by 3, if the sum of the digits is a multiple of 3, i.e. 21= 2 + 1=3, 36 = 3 + 6 = 9, 54 = 5 + 4 = 9.
3. A number with 3 or more digit is divisible by 4, if the number formed by its last two digit (i.e. ones and tens) is divisible by 4.
For example, in 212, number formed by ones and tens digit is 12 which is divisible by 4. So, 212 is divisible by 4.
In 1938, number formed by ones and tens digit is 38 which is not divisible by 4. Hence, 1938 is not divisible by 4.
4. A number is divisible by 5, if it has either 0 or 5 at its ones place.(Examples: 5, 10, 15, 20, 25, .......)
5. A number is divisible by 6, if it is divisible by 2 and 3 both. (Examples: 12, 18, 24, .......)
6. A number with 4 or more digit is divisible by 8, if the number formed by the last three digits is divisible by 8. Example, in 9216, number formed by the last three digits is 216, which is divisible by 8. Hence, 9216 is divisible by 8.
7. A number is divisible by 9, if the sum of the digits of the number is divisible by 9. Examples: 1+8 = 9, 2+7= 9, 4+5= 9.
8. Divisibility by 10: if a number has 0 at its ones place then it is divisible by 10. (Examples: 10, 20, 30, 40, 50,…….)
9. Divisibility by 11: 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.
For example, 61809, sum of the digit at odd places= 6+8+9= 23 and sum of the digit at even places = 1+0= 1. Difference = 23 – 1 = 22, is divisible by 11.
Thus, 61809 is divisible by 11.