Find the smallest number consisting of only "1"-s,
which would be divisible by "333...3" (one hundred "3"-s.)
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Posted 09 February 2013 - 09:36 PM Best Answer
Posted 10 February 2013 - 12:44 AM
Spoiler for answer3333....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 by Prime, 10 February 2013 - 12:44 AM.
Past prime, actually.
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