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