I'll prove this by induction. For \(n=1\) it's obvious, as the digit sum is divisible by \(3\). Now, let's asume the statement is true for some number with \(3^k\) digits. The corresponding \(3^{k+1}\) number is formed by multiplying the \(3^k\) digit number by \(100...00100...001\): a 1 followed by \(3^k-1\) zeros, followed by a 1, followed by \(3^k-1\) zeros, followed by a 1. This number has a digit sum of 3, so it is divisible by 3. Therefore, the new \(3^{k+1}\) digit number is divisible by \(3^{k+1}\). Thus, the statement is correct for all natural numbers \(n\).

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TopNewestI'll prove this by induction. For \(n=1\) it's obvious, as the digit sum is divisible by \(3\). Now, let's asume the statement is true for some number with \(3^k\) digits. The corresponding \(3^{k+1}\) number is formed by multiplying the \(3^k\) digit number by \(100...00100...001\): a 1 followed by \(3^k-1\) zeros, followed by a 1, followed by \(3^k-1\) zeros, followed by a 1. This number has a digit sum of 3, so it is divisible by 3. Therefore, the new \(3^{k+1}\) digit number is divisible by \(3^{k+1}\). Thus, the statement is correct for all natural numbers \(n\).

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