How do you find the divisibility rule for 3?

Answer

Take the sum of the digits of the number. If that sum is a multiple of 3 then the number is divisible by 3. For example 456, the sum of the digits is 15 which 1+5=6, which is a multiple of 3. Therefore it is divisble by 3. In fact 456/3 = 152

Answer

Let's try to put it simply.

It is crucial yet simple to understand that a number "can be divided by Three" only if the rest of its division by Three is equal to Zero. This number is thus a multiple of Three.

Now, if I have a number N, lets call [N]₃ = N%3 = the rest of the division by 3.

For example:

[17]₃ = 2

because 17 = 5*3 + 2

[21]₃ = 0

because 21 = 7*3 + 0 ( this means 21 can be divided by 3 )

One can easily convince himself that :

[a+b]₃ = (a+b)%3 = a%3 + b%3 = [a]₃ + [b]₃

and that :

[a*b]₃ = (a*b)%3 = (a%3)*(b%3) = [a]₃*[b]₃

(Although this demands deeper proof, but lets keep it simple)

Basically, what this verifies is that you can take the "plus" an "times" signs in and out of the parenthesis.

We will also need to notice that :

[10]₃ = [9]₃ + [1]₃ = 0+1 = 1

Now, any number N can be wrote some other way by saying :

N = 10^k * A_k + 10^k-1 * A_k-1 + ... + 10 * A₁ + A₀

where A_m = digits ( for m = 1, 2, ..., k which means m going from 1 to k)

For example:

N = 56572 = 10⁴ * 5 + 10³ * 6 + 10² * 5 + 10¹ * 7 + 2

N = 56572 = 50000 + 6000 + 500 + 70 + 2



Now don't get scared, this seems WAY more complicated than it is, we're going to look at [N]₃ again.

[N]₃ = [ 10^k * A_k + 10^k-1 * A_k-1 + ... + 10 * A₁ + A₀ ]₃

Here I just use the equality above for N.

But now, from what we saw just earlier, we can get the + and the * out of the [ ]₃ so :

[N]₃ = ([10]₃)^k * [A_k]₃ + ([10]₃)^k-1 * [A_k-1]₃ + ... + [10]₃ * [A₁]₃ + [A₀]₃

and since we know [10]₃ = 1 we have :

[N]₃ = [A_k]₃ + [A_k-1]₃ + ... + [A₁]₃ + [A₀]₃

because ([10]₃ )^k = (1)^k = 1 and 1*A_m = A_m

and finally we put the + back in the [ ]₃

[N]₃ = [ A_k + A_k-1 + ... + A₁ + A₀ ]₃

Which is the rest of the division by Three of the digits sum.

So the rest of the division of the digits sum equals the rest of the division of the original number N !!!

If one equals Zero, the other must equal Zero too.

As a result, if you want to know if a number can be divided by Three ( if the rest of its division by Three equals Zero ), you just need to know if the sum of its digits can be divided by Three, that is, if this sum turns out to be a multiple of Three.

Comments welcomed.

First answer by Scienceguy22. Last edit by SeldomSeen77. Contributor trust: 14 [recommend contributor]. Question popularity: 33 [recommend question]

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