Why is the divisibility rule of 3?

 

The divisibility rule of 3 is simple:

For a given number, add the individual digits. If the sum is divisible by 3, then the overall number is divisible by 3. Otherwise, it's not.

But why the sum of the digits?

How can we prove that this is robust for any given number?

If you wanna think it out yourself, take a pause. Otherwise, just continue reading below...

The Proof:

Consider a generalized decimal number, with the total number of digits as n+1, and the coefficients being a₀, a₁, a₂, and so on up to a_n. So, our number can be expressed as:

a₀10⁰ + a₁10¹ + a₂10² + ... + a_n10ⁿ.

Now, the powers of 10 can be written in terms of 9s, right?

10¹ = 10 = 9 + 1

10² = 100 = 99 + 1

10³ = 1000 = 999 + 1 and similarly,

10ⁿ = 100..0 = 99..9 + 1.

Then, our number becomes:

a₀ + a₁(9 + 1) + a₂(99 + 1) + .... + a_n(99..9 + 1)

= [a₀ + a₁ + a₂ + ... + a_n] + [9*a₁ + 99*a₂ + .... + 99..9*a_n]

= [a₀ + a₁ + ... +a_n]+ [3*(3*a₁ + 33*a₂ + .... + 33..3*a_n)]

Yeah, we have two groups: 

One group has the sum of digits; the other group is already a multiple of 3, independent of the given number!

If we want the overall number to be divisible by 3, just we need to make sure that the first group is divisible by 3, i.e., the sum of digits of the number should be divisible by 3! Ta-daa!

So, when someone asks you about "why is the rule", you can say that's why!

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P. S. Thanks for reading! Feel free to share your thoughts in the comments section...