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

- Ask someone else, say
**B**, to take*p*times as many, where*p*is any number you like to choose. - Request
**A**to give*q*of his counters to**B**, where*q*is any number you like to select. - 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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