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

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 use the concept of modular arithmetic.

Let’s consider a number represented as d_nd_n-1…d₁d₀, where each d represents a digit of the number. This can be expressed in expanded form as:

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

Now, we know that each power of 10 can be expressed in terms of modulo 3:

10 ≡ 1 (mod 3)

This equivalence indicates that:

10^k ≡ 1 (mod 3) for any integer k

Using this property, we can rewrite our number N modulo 3:

N ≡ d_n imes 1 + d_{n-1} imes 1 + … + d_1 imes 1 + d_0 imes 1 (mod 3)

Thus, we simplify to:

N ≡ d_n + d_{n-1} + … + d_1 + d_0 (mod 3)

This shows that the remainder of N when divided by 3 is the same as the remainder of the sum of its digits when divided by 3.

Now, if the sum of the digits S = d_n + d_{n-1} + … + d_1 + d_0 is divisible by 3, it means:

S ≡ 0 (mod 3)

So, since N ≡ S (mod 3), we have:

N ≡ 0 (mod 3)

This implies that N is also divisible by 3.

In conclusion, we have proved that if the sum of the digits of a number is divisible by 3, then the number itself is indeed divisible by 3.