Yes, a number is divisible by 3 if and only if the sum of its digits is divisible by 3. This is a well-established divisibility rule.
Understanding the Divisibility Rule of 3
This rule works because of the properties of base-10 arithmetic. Any number can be expressed as a sum of its digits multiplied by powers of 10. For example, the number 123 can be written as (1 x 100) + (2 x 10) + (3 x 1).
Since 10 ≡ 1 (mod 3), any power of 10 is also congruent to 1 (mod 3). Therefore, when we find the remainder when a number is divided by 3, we only need to consider the sum of its digits. If this sum is divisible by 3, then the original number is also divisible by 3.
- Example 1: Consider the number 27. The sum of its digits is 2 + 7 = 9, which is divisible by 3. Therefore, 27 is divisible by 3.
- Example 2: Consider the number 459. The sum of its digits is 4 + 5 + 9 = 18, which is divisible by 3. Therefore, 459 is divisible by 3.
- Example 3: Consider the number 29. The sum of its digits is 2 + 9 = 11, which is not divisible by 3. Therefore, 29 is not divisible by 3.
Multiple sources (Byjus.com, Cuemath, Reddit, StackExchange) confirm this divisibility rule and provide various explanations and proofs. The rule's effectiveness stems from the relationship between powers of 10 and the modulus 3.
The core concept is that congruence modulo 3 simplifies the divisibility check. The proof often involves representing the number in expanded form and showing that the remainder when divided by 3 depends solely on the sum of the digits.
