TargetIIT

here's a simple one

prove that a number with 3^m equal digits is divisible by 3^m

Induction is a dependable friend

For m =1, any such number is a multiple of 111 which is divisible by 3

Let us assume the truth of the statement for m = k.

i.e. the number N[ss]k[/ss] = nnn....nnn (3[p]k[/p] times) is divisible by 3[p]k[/p]

Now the number N[ss]k+1[/ss] = nnn....nnn (3[p]k+1[/p] times) consists of 3 blocks of the number N[ss]k[/ss]

so N[ss]k+1[/ss] = N[ss]k[/ss] [1+10[p][p]3[p]k[/p][/p][/p]+10[p]2.3[p]k[/p][/p]]

The bracketed term is divisible by 3 and N[ss]k[/ss] is divisible by 3[p]k[/p] and hence N[ss]k+1[/ss] is divisible by 3[p]k+1[/p] which wraps up the induction process

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