Who Has It?

This clever trick requires three little things that can be put in one’s pocket – a pencil, a key, and a penknife will do very well. In addition to that, place a plate with 24 nuts – draughts or domino pieces or matches will do just as well – on the table.

Having completed these preparations, ask each of your three friends to put one of the three things into his pocket – one the pencil, the second the key, and the third the penknife. This they must do in your absence, and when you return to the room, you guess correctly where each object is.

The process of guessing is as follows: on your return (i.e., after each has concealed the object) you ask your friends to take care of some nuts – you give one nut to the first, two to the second, and three to the third. Then you leave the room again, telling them that they must take more nuts – the one who has the pencil should take as many nuts as he was given the first time; the one with the key twice as many as he has been given; and the one with the penknife four times the number. The rest, you tell them, should remain in the plate.

When they have done that, they call you into the room. You walk in, look at the plate and announce what each of your friends has in his pocket.

The trick is all the more mystifying since you do it solo, so to speak, without any assistant who could signal to you secretly.

There is really nothing tricky in the riddle – the whole thing is based on calculation. You guess who has each of the things from the number of nuts remaining in the plate. Usually there are not many of them left – from one to seven – and you can count them at once glance.

How, then, do you know who has what thing?

Simple. Each different distribution of the three objects leaves a different number of nuts in the plate. Here is how it is done.

Let us call your three friends Dan, Ed, and Frank, or simply D, E, F. The three things will be as follows: the pencil – a, the key – b, and the penknife – c. The three objects can be distributed among the trio in just six ways:

D E F
-----
a b c
a c b
b a c
b c a
c a b
c b a

There can be no other combinations – the table above exhausts all of them.

Now let us see how many nuts remain after each combination:

DEF    number of nuts taken         total remainder
abc     1+1=2; 2+4=6; 3+12=15    23     1
acb     1+1=2; 2+8=10; 3+6=9      21     3
bac     1+2=3; 2+2=4; 3+12=15    22     2
bca     1+2=3; 2+8=10; 3+3=6      19     5
cab     1+4=5; 2+2=4; 3+6=9        18     6
cba     1+4=5; 2+4=6; 3+3=6        17     7

You will see that the remainder is each time different. Knowing what it is, you will easily establish who has what in his pocket. Once again, for the third time, you leave the room, look at your notebook into which you have written the table above (frankly speaking, you need only the first and last columns). It is difficult to memorize this table, but then there is really no need for that. The table will tell where each thing is.

If, for example, there are five nuts remaining in the plate, the combination is bca, i.e.,
Dan has the key,
Ed has the penknife, and
Frank has the pencil.

If you want to succeed, you must remember how many nuts you gave each of your three friends (the best way is to do so in alphabetic order, as we have done here).