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# very interesting problem!

Prove that for every number consisting of $$3^{n}$$ equal digits is divisible by $$3^{n}$$.

Note by Kiran Patel
3 years, 9 months ago

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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 $$n$$. · 3 years, 9 months ago