Three real numbers a, b, c with a < b < c, are said to be in arithmetic progression if c - b = b - a.

Define a sequence u(n), n = 0, 1, 2, 3, ... as follows: u(0) = 0, u(1) = 1 and, for each n > 0, u(n+1) is the smallest positive integer such that u(n+1) > u(n) and {u(0), u(1),... u(n), u(n+1)} contains no three elements that are in arithmetic progression.

Find u(100).

[Irish MO, Paper 1, Question 5, 1999]

**Feel free to comment, ask questions and even check your answer in the comments box below powered by**~~Disqus~~ Google+.

}

This space is here to avoid seeing the answers before trying the problem!

}

If you enjoy using this website then please consider making a donation - every little helps :-)

You can receive these questions directly to your email box or read them in an RSS reader. Subscribe using the links on the right.

Don’t forget to follow Gifted Mathematics on Google+, Facebook or Twitter. You may add your own interesting questions on our Google+ Community and Facebook..

You can also subscribe to our Bookmarks on StumbleUpon and Pinterest. Many resources never make it onto the pages of Gifted Mathematics but are stored in these bookmarking websites to share with you.}

This space is here to avoid seeing the answers before trying the problem!

}

If you enjoy using this website then please consider making a donation - every little helps :-)

You can receive these questions directly to your email box or read them in an RSS reader. Subscribe using the links on the right.

Don’t forget to follow Gifted Mathematics on Google+, Facebook or Twitter. You may add your own interesting questions on our Google+ Community and Facebook..

You can also subscribe to our Bookmarks on StumbleUpon and Pinterest. Many resources never make it onto the pages of Gifted Mathematics but are stored in these bookmarking websites to share with you.

## No comments:

## Post a comment