How can we prove that if the sum of a number’s digits is divisible by 3, then the number itself is also divisible by 3?

by

To prove that if the sum of the digits of a number is divisible by 3, then the number itself is also divisible by 3, we can utilize modular arithmetic and the properties of numbers in base 10.

Let’s denote a number as N. We can express N in terms of its digits. For a number with digits dn, dn-1, …, d1, d0 (where dn is the most significant digit and d0 is the least significant), we can write:

N = d_n 	imes 10^n + d_{n-1} 	imes 10^{n-1} + ... + d_1 	imes 10^1 + d_0 	imes 10^0

Now, to see how the number behaves under modulo 3, we can look at the powers of 10:

  • 10 mod 3 = 1
  • 10^2 mod 3 = 1
  • 10^3 mod 3 = 1
  • …

Since all powers of 10 are congruent to 1 modulo 3, we can rewrite our number N under modulo 3 as:

N mod 3 = (d_n + d_{n-1} + ... + d_1 + d_0) mod 3

This means that N mod 3 is equal to the sum of its digits modulo 3. Therefore, if the sum of the digits (let’s denote it as S) is divisible by 3, it follows that:

S mod 3 = 0

As a result, we can conclude:

N mod 3 = 0

This demonstrates that N is also divisible by 3. Thus, we have shown that if the sum of the digits of a number is divisible by 3, the number itself must also be divisible by 3.

In summary, the divisibility rule for 3 hinges on the consistent behavior of the digits of a number when expressed in base 10 and examined under modulo 3. This remarkable property simplifies the checking of divisibility by merely summing the digits.

Leave a Comment