Divisibility Rules: 0-10

A list of divisibility tests for factoring integers.

0 – No number is ever divisible by zero.  This is undefined.

1 – All numbers are divisible by one. Nothing is achieved by doing so. Completely trivial.

2 – Only even numbers are divisible by two.  That is, any number that ends in a 0, 2, 4, 6, or an 8.

3 – If the sum of the digits of the integer is divisible by three, then so is the integer itself.  This is iterative – you may find the sum of the digits of the sum of the digits of the sum of the digits, so on so forth, until you are left with a single digit number: 0, 3, 6, or 9, (or any larger number that is also divisible by 3).

4 – If the number is divisible by 2 twice then it is divisible by 4.  That is, if after factoring out a 2 you can factor out another 2 from the remaining factor then the original integer is a multiple of 4.

4 – Alternatively, take the last two digits (the least significant digits – the ones and the tens place).  If these two digits are themselves divisible by 4 then the whole integer is divisible by 4.

5 – All numbers ending in a 0 or a 5 in the ones place are divisible by 5.

6 – The prime factors of 6 are 2 and 3.  Thus, any number that is divisible by both 2 and 3, simultaneously, is also divisible by 6.  It must pass both tests – refer to the tests for divisibility by 2 and by 3.  If the integer’s digits add to a multiple of 3 and the integer is even, then its a multiple of six.

7 – Suppose the integer s is in the form , where m and n are both integers as well, but restrict n to a single digit.  This is saying, take n to be the ones digit and take m to be everything to the left of the ones digit as its own integer.  For example, , therefore and .  The process is relatively simple.  Double n and subtract it from m: .  In this example we take 36 and we subtract 8 for a difference of 28.  Now, 364 is divisiby by 7 only if 28 is divisible by 7.  That is to say, more generally, is divisiby by 7 if is divisible by 7.  The process is reiterative.

7 – An alternative method exists.  Memorize the set .  What we want to do is pair each digit of your number with a digit from the set of six values, above.  The ones digit gets paired with the first value, the tens digit gets paired with the second value, the hundreds digit gets paired with the third value, etcetera.  When you have reached the end of the set, cycle around to the beginning.  Continue pairing digits of the number with a value from the set until all digits have been paired.  Multiply these pairings and add their products.  For example, take the number .  We add the products of the associated pairs:

.

The value 2037 is divisible by 7 only if 28 is divisible by 7.  The algorithm is reiterative.

8 – This divisibility test for 8 is not unlike the divisibility test for 4.  A number must be divisible by 2 three distinct times.  . That is, after dividing by 2 evenly, the remaining factor must still be divisibly by 4.

8 – The alternative divisibility test for 8 is not unlike the alternative divisibility test for 4.  We take the three least significant digits (the ones, tens and hundreds place).  As its own unique integer, if it is divisible by 8 then the whole is divisible by 8.

9 – The divisibility test for 9 is not unlike the divisibility test for 3.  We must add up all the digits of the integer, re-iteratively if you wish.  The sum must ultimately be divisible by 9.

9 – Alternatively, not unlike the divisibility test for four, 9 is the square of 3.  So a number is divisible by 9 if it is divisible by 3, and the remaining factor is also divisible by 3.

10 – If the integer ends in a zero then it is divisible by 10.  This is the same as passing the divisibility tests for 2 and 5 simultaneously.  For 5, the number must end in a 0 or 5, while for 2 the number must be even.

For divisibility rules for integers larger than ten, refer to this followup post.

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