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