Prime Posted February 9, 2013 Report Share Posted February 9, 2013 Find the smallest number consisting of only "1"-s, which would be divisible by "333...3" (one hundred "3"-s.) (Decimal system.) Quote Link to comment Share on other sites More sharing options...
0 googon97 Posted February 9, 2013 Report Share Posted February 9, 2013 3333....3 is 3*1111....1 The number consisting of only ones must be a multiple of 3 and 1111...1. For it to be a multiple of 111....1, it must be a multiple of 100 1's long. The first multiple of 100 divisible by 3 is 300. So, the number is 111...1 with 300 1's. 1 Quote Link to comment Share on other sites More sharing options...
0 Prime Posted February 9, 2013 Author Report Share Posted February 9, 2013 (edited) 3333....3 is 3*1111....1 The number consisting of only ones must be a multiple of 3 and 1111...1. For it to be a multiple of 111....1, it must be a multiple of 100 1's long. The first multiple of 100 divisible by 3 is 300. So, the number is 111...1 with 300 1's. That's the number I was looking for! Edited February 9, 2013 by Prime Quote Link to comment Share on other sites More sharing options...
Question
Prime
Find the smallest number consisting of only "1"-s,
which would be divisible by "333...3" (one hundred "3"-s.)
(Decimal system.)
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