The discussion focuses on proving that a positive integer \( n \in \mathbb{Z}^+ \) is divisible by 3 (and 9) if and only if the sum of its digits is divisible by 3. The proof involves demonstrating that \( n \equiv 0 \mod 3 \) can be derived from the expression \( n = X_1 \cdot 10^n + X_2 \cdot 10^{n-1} + \ldots + X_n \), leading to the conclusion that \( (X_1 + X_2 + \ldots + X_n) \equiv 0 \mod 3 \). Additionally, the discussion references the property \( 10^k \equiv 1 \mod 3 \) and suggests researching "casting out nines" for further insights.
PREREQUISITESMathematics students, educators, and anyone interested in number theory, particularly those studying divisibility rules and modular arithmetic.
yeland404 said:Homework Statement
Prove that n ℂ Z+ is divisible by 3( respectively 9). to show that if and only if the sum of its digits is divisible by 3
Homework Equations
The Attempt at a Solution
so n= 3q, q>3 that n\equiv0 mod 3
n=X1* 10^n+ x2*10^n-1...Xn
so need to prove(x1+x2+...Xn)\equiv0 mod 3, the how to prove the next step