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 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

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 Solution | Comment 1 of 2
Ignoring y
x             floor (Log2(1/x) )
1/2 to 1        0
1/4 to 1/2     1
1/8 to 1/4     2
1/16 to 1/8   3
1/32 to 1/16 4

The sum of the widths of the intervals for which x  floor (Log2(1/x) ) is even is 1/2 + 1/8 + 1/32 + ... = 2/3

Since we want both x and y to be this way the area is (2/3)*(2/3)=4/9

But x<y only half of the time so (4/9)/2 = 2/9.
P/Q = 2/9

I didn't notice the solution had been boiled down to a single number when this was in the queue or I would have suggesting changing the problem to just finding the area.

But the solution to the stated problem is
P+Q = 11

 Posted by Jer on 2016-02-15 12:30:34

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