Divisibility rules in odia||ବିଭାଜ୍ୟତା ସୂତ୍ର||divisibility rules for 1,2,3,4,5,6,7 upto 29 in odia

#divisibility_rules
ବିଭାଜ୍ୟତା ସୂତ୍ର
Divisibility rules in odia
Divisibility rules for 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29

Divisibility rule for 1
Every number is divisible by 1.
Example: 5 is divisible by 1
Divisibility rule for 2
Any even number or number whose last digit is an even number (0, 2, 4, 6, 8) is divisible by
Example: 220 is divisible by 2.

Divisibility rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.
Example: 315 is divisible by 3.
Here, 3 + 1 + 5 = 9

9 is divisible by 3. It means 315 is also divisible by 3.

Divisibility rule for 4
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
Example: Example: 7568 is divisible by 4
Here, 68 is divisible by 4 (68÷4 = 17)
Therefore, 7568 is divisible by 4

Divisibility rule for 5
A number is exactly divisible by 5 if it has the digits 0 or 5 at one’s place.
Example: 5900, 57895, 4400, 1010 are divisible by 5.
Divisibility rule for 6
A number is exactly divisible by 6 if that number is divisible by 2 and 3 both. It is because 2 and 3 are prime factors of 6.
Example: 63894 is divisible by 6, the last digit is 4, so divisible by 2, and sum 6+3+8+9+4 = 30 is divisible by 3.
Divisibility rule for 7
Double the last digit and subtract it from the remaining leading truncated number to check if the result is divisible by 7 until no further division is possible
Example: 1093 is divisible by 7
Remove 3 from the number and double it = 6

Remaining number is 109, now subtract 6 from 109 = 109 – 6 = 103.

Repeat the process, We have last digit as 3, double = 6

Remaining number is 10, now subtract 6 from 10 = 10 – 6 = 4.

As 4 is not divisible by 7, hence the number 1093 is not divisible by 7