What are the divisibility rules from 2-12?

Answer 1

14

Answer 2

• 2 - the last (units) digit is divisible by 2 (ie the number is even);
• 3 - the sum of its digits is divisible by 3
• 4 - the last two digits are divisible by 4; or

the sum of the last (units) digit and twice the tens digit is divisible by 4;

• 5 - the last (units) digit of the number is 0 or 5;
• 6 - the rules for both 2 and 3 are true:

the last (units) digit is divisible by 2 (number is even), and

the sum of its digits is divisible by 3;

• 7 - see below
• 8 - the last three digits are divisible by 8; or

the sum of the last (units) digit, twice the tens digit and four times the hundreds digit is divisible by 8;

• 9 - the sum of its digits is divisible by 9;
• 10 - the last (units) digit is 0;
• 11 - the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 11, or is 0 (the sums are equal);
• 12 - the rules for both 3 and 4 are true:

the sum of its digits is divisible by 3, and

the last two digits are divisible by 4 or the sum of the last (units) digit and twice the tens digit is divisible by 4;

Where a sum is required to be divisible, the sum can be tested in the same way; repeating the rule until a single digit remains makes it easy to check.

Where there are 2 alternative rules (for 4 and 8) I've given the "standard" rule (ie the one I was taught) first followed by an alternative rule I've since discovered that is much easier.

The rule for 7:

There is no simple rule for divisibility by 7 and most of the invented rules often take just as long as (if not longer than) dividing the original number by 7.

The best I can offer for 7:

1. split the number into blocks of 3 digits starting at the right hand end (the units digit) - just like when reading the number;
2. in each block add twice the first digit to thrice the second digit to the third digit (ie in each block it is 2 x hundreds digit + 3 x tens digit + units digit);
3. alternately subtract and add the sums of the blocks;
4. if this sum is 0 or divisible by 7, the original number is divisible by 7.

for example: is 5678534 divisible by 7?

5678534 → 5,678,534

→ [0x2 + 0x3 + 5] - [6x2 + 7x3 + 8] + [5x2 + 3x3 + 4]

→ 5 - 41 + 23 = -13 which is not divisible by 7, so 5678534 is not divisible by 7.

The rule for 11 is also expressed as:

Alternately subtract and add the digits from the right hand end of the number; if the result is 0 or divisible by 11, the original number is divisible by 11.