Following is an excerpt from
W.W.Rouse Ball and H.S.M.Coxeter,
Mathematical Recreations and Essays

No Questions Asked

Ask A to take any number of counters that he pleases: suppose that he takes n > 1 counters.

  1. Ask someone else, say B, to take p times as many, where p is any number you like to choose.
  2. Request A to give q of his counters to B, where q is any number you like to select.
  3. Next, ask B to transfer to A a number of counters equal to p times as many counters as A has in his possession. Then there will remain in B's hands q(p + 1) counters; this number is known to you; and the trick can be finished either by mentioning it or in any other way you like.

The reason is as follows. The result of operation (ii) is that B has pn + q counters, and A has n - q counters. The result of (iii) is that B transfers p(n - q) counters to A; hence he has left in his possession (pn + q) - p(n - q) counters, that is, he has q(p + 1) .

For example, if originally A took any number of counters, then (if you chose p equal to 2), first you would ask B to take twice as many counters as A had done; next (if you chose q equal to 3) you would ask A to give 3 counters to B; and then you would ask B to give to A a number of counters equal to twice the number then in A's possession; after this was done you would know that B had 3(2 + 1), that is, 9 left.

This trick (as also some of the following problems) may be performed equally well with one person, in which case A may stand for his right hand and B for his left hand.

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