There 100 people standing in a circle who are numbered 1 to 100 in order, with 1 standing between 100 and 2.
Person 1 is handed a gun, shoots Person 2 and hands the gun to Person 3. Person 3 shoots Person 4 and hands the gun to Person 5. This continues round and round the circle, with each person who receives the gun shooting the person on their left and then handing the gun to the person on the left of the person they have just shot.
Assuming the gun never runs out of bullets, what number person survives?
On the first circuit of the circle the odd-numbered people shoot the even-numbered people. That is 2x+1 where x is 0 to 49 survive.
On the second circuit people numbered 4x+1 where x is 0 to 24 survive (1, 5, 9 . . . 97).
On the third circuit it is a little more complicated as there are 25 people in the circle which is an odd number. Initially people numbered 8x+1 where x is 0 to 12 survive (1, 9 . . . 97), but 97 shoots 1, leaving people numbered 8x+1 where x is 1 to 12 (9, 17 . . . 97).
On the fourth circuit there are 12 people in the circle, and people numbered 16x+9 where x is 0 to 5 survive (9, 25, 41, 57, 73 and 89).
On the fifth circuit there are six people in the circle, and people numbered 32x+9 where x is 0 to 2 survive (9, 41 and 73).
And now it is clear that 9 shoots 41 and 73 shoots 9, leaving 73 as the survivor.