Question

What are are the tests of divisibility of various numbers?

Answer 1

There are many divisibility tests. Here are a few, along with how they can be derived.

• An integer is divisible by `2` if the final digit is even.

• An integer is divisible by `3` if the sum of its digits is divisible by 3.

• An integer is divisible by `4` if the integer formed by the last two digits is divisible by 4.

• An integer is divisible by `5` if the final digit is 5 or 0.

• An integer is divisible by `6` if it is divisible by 2 and by 3.

• An integer is divisible by `7` if subtracting twice the last digit from the integer formed by removing the last digit is a multiple of 7.

• An integer is divisible by `8` if the integer formed by the last three digits is divisible by 8 (this can be made easier by noting that the rule is the same as for 4s if the hundreds digit is even, and the opposite otherwise)

• An integer is divisible by `9` if the sum of the digits is divisible by 9.

• An integer is divisible by `10` if the last digit is `0`

For these and more, take a look at the wikipedia page for divisibility rules .

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Now, one may wonder about how to come up with these rules, or at least show that they actually will work. One way to do this is with a type of math called modular arithmetic .

In modular arithmetic, we pick an integer `n` as the modulus and then treat every other integer as being congruent modulo `n` to its remainder when divided by `n`. An easy way to think about this is that you can add or subtract `n` without changing the value of an integer modulo n. This is the same as how, on an analog clock, adding twelve hours results in the same time. Adding hours on a clock is addition modulo `12`.

What makes modular arithmetic very useful in determining divisibility rules is that for any integer `a` and positive integer `b`, we can say that `a` is divisible by `b` if and only if

`a-=0 " (mod b)"` (`a` is congruent to `0` modulo `b`).

Let's use this to see why the divisibility rule for `3` works. We'll do so using an example which should show the general concept. In this example, we'll see why `53412` is divisible by `3`. Remember that adding or subtracting `3` will not change the value of an integer modulo `3`.

`53412` is divisible by `3` if and only if `53412 -= 0 " (mod 3)"`

But also, because `10 -3 -3 -3 = 1`, we have `10 -= 1 " (mod 3)"`

Thus:
`53412 -= 5*10^5+3*10^4+4*10^3+1*10^2+2 " (mod 3)"`

`-= 5*1^5+3*1^4+4*1^3+1*1^2+2 " (mod 3)"`

`color(red)(-= 5+3+4+1+2 " (mod 3)")`

`-= 3*5 " (mod 3)"`

`-= 0*5 " (mod 3)"`

`-= 0 " (mod 3)"`

Thus `53412` is divisible by `3`. The step in red demonstrates why we can simply sum the digits and check that instead of trying to divide the original number by `3`.