A modified game of Russian Roulette.
100 Russians are standing in a circle in with numbers 1 to 100 written on their backs .
The first Russian (No 1) has a gun. He shoots and kills the next person (i.e. No. 2) and gives the gun to the next Russian alive (i.e. No. 3).
All of them do the same until only 1 survives.
Which number is the last to survive
The answer is 73.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Solution which is a modification of the Josephus problem
The Josephus problem (or Josephus permutation) is a theoretical problem related to a counting-out scenario.
This requires writing the nearest number to 100 that is the power of 2, in this case 64 and subtract it from 100 to get 36
Multiply this by 2 and add 1 to it to get the result.