# Divisibility Rules

Given a number $x$, can you determine if it is divisible by a number say $y$ with out even performing the actual division. Divisibility rules are quite handy when you want to test for divisibility of numbers even before you can actual divide the number.

For instance, to check if a number is divisible by 7, the general divisibility rule for 7 is to

1. Double the units digit of the number, and subtract it from the rest of the number.
2. If the answer obtained is divisible by 7, then the actual number is divisible by 7.
3. If you are not sure about the obtained answer’s divisibility, apply the rule again on the answer.

Given below is the table containing the divisibility rules for numbers from 2 to 13.

NumberDivisibility RuleExample
2Number must be even124 is even, hence it is divisible by 2.
421 is odd, hence it is not divisible by 2
3Sum of the digits must be divisible by 3Consider 123. The sum of its digits is 1+2+3 = 6. 6 is divisible by 3. Hence 123 is divisible by 3
4The last two digits must be divisible by 4In 123496, the last two digits 96 is divisible by 4, hence 123496 is divisible by 4.
5The last digit is either 5 or 0, then the number is divisible by 5 947245.
Here the last digit is 5. Hence 947245 is divisible by 5.
6The number must be divisible by both 2 and 396312.
The number is even hence its divisible by 2 and the sum of its digits is 21, which is divisible by 3. Hence 96312 is divisible by 6
7Multiply the last digit by 2, then subtract it from the rest of the number.
If the result is divisible by 7, then the number is divisible by 7
672.
1. Multiplying the last digit 2 by 2 gives 4.
2. Subtract 4 from the rest of the number i.e 67 - 4 = 63.
Since 63 is divisible by 7. 672 is divisible by 7
8The last three digits must be divisible by 826032.
Since the last three digits 032 is divisible by 8. The number 26032 is divisible by 8
9Sum of the digits must be divisible by 911106.
The sum of the digits is 1+1+1+0+6 = 9, which is divisible by 9. Hence 11106 is divisible by 9.
10If the last digit is 0, the number is divisible by 10123450.
Since the number ends in 0. It is divisible by 10
11Subtract the sum of the digits in even numbered places with the sum of the digits in odd numbered places. The result obtained must be divisible by 11107415.
Sum of the digits in even places is 1+7+1=9
Sum of the digits in odd places is 0+4+5 = 9
Difference in the even and odd places sum is 9-9=0, which is divisible by 11.
Hence 107415 is divisible by 11
12Number must be divisible by both 3 and 4132.
Sum of the digits is 1+3+2 = 6, which is divisible by 3. Hence the number is divisible by 3.
Last two digits 32 is divisible by 4. Hence the number is divisible by 4.
Since the number is divisible by 3 and 4. The number is divisible by 12.
13Multiply the last digit by 9, then subtract it from the rest of the number. If the result obtained is divisible by 13, then the number is divisible by 13 247.
Multiply 7 by 9, which gives 63.
Subtract 63 from 24, which gives -39.
-39 is divisible by 13. Hence 247 is divisible by 13.

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