Divisible by power of 3

Show that the number 11…1 with \(3^n \) digits is divisible by \(3^n\)

Solution:

We use induction. For n=1; we check 111 is divisible by 3. Assuming that the result host for n=k, we establish that it holds for n=k+1.

The number 111…111 (with \(3^{k+1} \) digits) can be written in 3 blocks each having \(3^{k} \) 1’s.  Hence we can write it as \({111…111} \times 10^{(2 \times 3^k)} + {111…111} \times 10^{(3^k)} + {111…111}\) where {111…111} denotes \(3^k \) 1’s.
Taking {111…111} common we have \((111…111)(10^{(2 \cdot 3^k)} + 10^{(3^k)} + 1)\). By induction (111…111) is divisible by \(3^k \) and we also have 3 dividing \((10^{(2 \cdot 3^k)} + 10^{(3^k)} +1 )\) as it’s sum of digits is 3 (it has only three 1’s and rest are 0) .
Hence (111…111) having \(3^{(k+1)} \) 1’s is divisible by \(3^{(k+1)} \).

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