Two circular disks each have 10 0s and 10th 1s spaced equally around their edges in different orders. Show that the disks can always be superimposed on top of each other so that at least 10 positions have the same digit.

The superimposition is possible in a way such that at least 10 positions will have same digit.

Explanation:

The solution to this question is possible by using the pigeonhole principle. which is given as Suppose that n+1 (or more) objects are put into n boxes. Then some box contains at least two objects.

Suppose one disk is smaller than the other, but both the disks have equal number of sectors and they both have 10 0s and 10 1s on the sectors. Now fix the larger disk and superimpose the smaller disk on that. There are 20 positions of the smaller disk such that each sector of the smaller disk is contained in the larger disk. As the larger disk has 20 sectors with 10 0s and 10 1s, thus each sector of the smaller disk will match digits in exactly 10 out of the possible positions. The total number of digits matches are given as 10 times the number of sectors on the smaller disk and equals 200. So the average is given as 200/20 = 10

Thus Two circular disks each have 10 0s and 10th 1s spaced equally around their edges in different orders are superimposed on top of each other so that at least 10 positions have the same digit.