Prove that if the sum of the digits is divisible by 3, the number is divisible by 3.

The divisibility test for 3 is very important since it helps us to check if a number is divisible by 3. This is more useful in the case of larger numbers. It is known that we check if the sum of the digits is divisible by 3 to check its divisibility. Now, let's check how this rule is proved.

Answer: We split the number in the form of power of 10's to prove the rule of divisibility of 3.

Let's understand it in detail. The explanation is given below.

Explanation:

First, let's split the number in the form of a power of 10s. Let's take an example of a 3 digit number, abc, where a is hundred's digit, b is ten's digit and c is unit's digits.

Therefore, abc = 10² × a + 10 × b + c

Hence, abc = (99 + 1) × a + (9 + 1) × b + c

We can rearrange the equation as:

Hence, abc = 99a + 9b + a + b + c

When we divide the number by 3 we get:

abc / 3 = 33a + 3b + (a + b + c) / 3

Hence, abc is divisible by 3 only when (a + b + c) is divisible by 3.

Hence, We need to split the number in the form of power of 10s to prove the rule of divisibility of 3.