Find the smallest number consisting of only "1"-s,
which would be divisible by "333...3" (one hundred "3"-s.)
Jump to content
|Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.
Of course, you can also enjoy our collection of amazing optical illusions and cool math games.
If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.
Thanks and enjoy the Den :-)
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.
0 members, 0 guests, 0 anonymous users
Community Forum Software by IP.Board 3.4.5
Licensed to: BrainDen