 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Area From Even Floor (Posted on 2016-02-15) S denotes the set of pairs (x,y) with x ≤ y, 0 < x ≤ 1 and 0 < y ≤ 1, and having the restriction that:
Each of floor (Log2(1/x) ) and floor (Log2(1/y) ) is even.

Find P+Q, given that that the area of the graph of S is P/Q, where P and Q are relatively prime positive integers.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Sans Integration (spoiler) Comment 2 of 2 | Let's just do one dimension first.

What is the length of the line segment such that 0 < x <= 1?

Well, 1/x is between 1 and 2, or between 4 and 8, or between 16 and 32, etc,
So 1/x is between 1 and 1/2, or between 1/4 and 1/8, or between 1/16 and 1/32, etc.

Total length = 1/2 + 1/8 + 1/32 + ... = 2/3

Then the total area would be (2/3)^2 = 4/9 if x and y were independent.

However, x <= y, so one half the area qualifies.

Area = 2/9 = P/Q, so P+Q = 11

 Posted by Steve Herman on 2016-02-15 12:40:05 Please log in:
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