Problem:
Let S be the set of ordered pairs (x,y) such that 0<x≤1,0<y≤1, and ⌊log2​(x1​)⌋ and ⌊log5​(y1​)⌋ are both even. Given that the area of the graph of S is m/n, where m and n are relatively prime positive integers, find m+n. The notation ⌊z⌋ denotes the greatest integer that is less than or equal to z.
Solution:
Because ⌊log2​(x1​)⌋=2k for nonnegative integers k, conclude that 2k≤log2​(x1​)< 2k+1, so
22k≤x1​<22k+1, and 22k+11​<x≤22k1​
Similarly, for nonnegative integers k,
52k+11​<y≤52k1​
The graph consists of the intersection of two sets of rectangles. The rectangles in one set have vertical sides of length 1 and horizontal sides of lengths (1− 21​),(41​−81​),(161​−321​),…, and the rectangles in the other set have horizontal sides of length 1 and vertical sides of lengths (1−51​),(251​−1251​),(6251​−31251​),…. The intersection of the two sets of rectangles is also a set of rectangles whose total area is
[(1−21​)+(41​−81​)+⋯]⋅[(1−51​)+(251​−1251​)+⋯]=32​⋅65​=95​
so m+n=14​.
The problems on this page are the property of the MAA's American Mathematics Competitions