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.
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.