Prove that for any given positive integer n,there exists a number having digits 0,1 which is divisible by n.
Let that number be of the general form: x=¯bkbk−1...b1b0 where bi∈{0,1}.How can we construct x to be divisible by the given n?
Answer
Let ak=111...11⏟k=10k−19
So, among a1,a2,...an+1 two must have the same remainder modulo n, say ai and aj and i>j. Thus n∣ai−aj=10i−10j9=10j10i−j−19=111...11⏟i−j000...00⏟j
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