the second, and three to the third. Place the remaining 18 on the table, and then retire, that the three persons may distribute among themselves the things proposed, without your observing them. When the distribution has been made, desire the person who has the ring to take from the 18 remaining counters, as many as he has already; the one who has the shilling to take twice as many as he has already; and the person who has the glove to take four times as many. By these different combinations, the counters left can be only 1, 2, 3, 5, 6, or 7. When this is done, you may return, and by the number of counters left, you can discover what thing each has got, by employing the following words:

To make use of these words, you must recollect what has been already said, viz. that the number of the counters which remain, can be only 1, 2, 3, 5, 6 or 7, and never 4: you must observe also, that each syllable contains one of the vowels which we have made to represent the three things proposed, and that the above line must be considered as consisting only of six words: the first syllable of each word must also be supposed to represent the first person, and the second syllable the second person. This being comprehended, if there remains only one counter you must employ the first word, or rather the two first syllables par fer, the first of which, that containing A, shews that the first person has the ring represented by A; and the second syllable, that containing E, shews that the second person has the shilling, represented by E; from which you may easily conclude, that the third person has the glove. If two counters remain, you must take the second word César, the first syllable of which, containing E, will shew that the first person has the shilling, represented by E; and the second syllable, containing A, will indicate that the second person has

the ring, represented by A; you may then easily conclude that the third person has the glove.

PROBLEM XIX.

To tell, by inspecting a watch, at what hour a person has resolved to rise next morning.

1st. When the person has thought of an hour, bid him touch some other hour on the dial-plate, and then desire him to add 12 to it privately in his own mind, which will form a certain number.

2nd. Then desire him to proceed backwards, and to count the above number, beginning with the hour which he thought of.

Let the hour thought of, for example, be 8, and that touched be 3; as 12 added to 3 makes 15, desire the person to count that number, in a retrograde order from the hour touched, beginning with 8, the hour thought of; counting 8 on the hour 3, 9 on 2, 10 on 1, and so on, by which means 15 will fall upon the hour of 8.

The person will be surprised to find that he has fallen on the hour he thought of.

Two persons agree to take alternately numbers less than a given number, for example 11, and to add them together till one of them has reached a certain sum, such as 100; by what means can one of them infallibly attain to that number before the other?

The whole artifice of this problem consists in immediately making choice of the numbers 1, 12, 23, 34, and so on, or of a series which continually increases by 11, up to 100.

Let us suppose, that the first person, who knows the game, makes choice of 1; it is evident that his adversary, as he must count less than 11, can at most reach 11 by adding 10 to it. The first will then take 1, which will make 12; and whatever number the se

cond may add, the first will certainly win, provided he continually adds the number which forms the complement of that of his adversary to 11; that is to say, if the latter takes 8, he must take 3; if 9, he must take 2, and so on. By following this method, he will infallibly attain to 89; and it will then be impossible for the second to prevent him from getting first to 100; for whatever number the second takes, he can attain only to 99; after which the first may say, And 1 makes 100. If the second takes 1, after 89, it would make 90; and his adversary would finish by saying, And 10 make 100.

It is evident, that when two persons are equally well acquainted with the game, he who begins must necessarily win.

PROBLEM XXI.

Sixteen counters being disposed in two rows, to find that which a person has thought of.

The counters being arranged as follow, desire the person to think of one, and to observe well in which row it is:

Let us

suppose that the counter thought of is in the row A take up the whole row in the order in which it now stands, and dispose it in two rows C and D, in such a manner, that the first counter of the row A may be the first of the row C; the second of the row A, the first of the row D; and so on, transferring the 16

counters from A and B, to C and D. This being done, again ask in which of the vertical rows the counter thought of stands. We shall suppose it to be in C; remove that row as well as D, observing the same method as before; and continue in this manner until the counter thought of becomes the first of the row 1. If you then ask in which row it is, it may be immediately known, because after the last operation it will be the first in the row said to contain it; and as each row has a distinguishing character or sign, you may cause them all to be mixed with each other, and still be able to discover it by the sign you have remarked.

Instead of 16 counters, 16 cards may be employed. After you have discovered the one thought of, you may cause them to be mixed, which will conceal the artifice.

If a greater number of counters or cards be employed, disposed in two vertical rows, the counter or card thought of will not be at the top of the row after the last transposition: if there are 33 counters or cards, 4 transpositions will be necessary; if 64, there must be 5; and so on.

PROBLEM XXII.

A certain number of cards being shown to a person, to guess that which he has thought of.

To perform this trick, the number of the cards must be divisible by 3; and to do it with more convenience, the number must be odd.

The first condition, at least, being supposed, the cards must be disposed in three heaps with their faces turned upwards. Having then asked the person in which heap is the card thought of, place the heaps one above the other, in such a manner that the one containing the card thought of may be in the middle. Arrange the cards again in three heaps, and having asked in which of them is the card thought of, repeat the operation as before. Arrange them a third time in three heaps, and having once more asked the same

question, form them all into one heap, that containing the card thought of being in the middle. The card thought of must then necessarily be the middle one, that is to say, if 15 cards have been employed, it will be the eighth from the top; if 21, the eleventh; if 27, the fourteenth; and so on. When the number of the cards is 24, it will be the twelfth, &c.

PROBLEM XXIII.

To arrange 30 criminals in such a manner, that by counting them in succession, always beginning again at the first, and rejecting every ninth person, 15 of them may be saved.

Arrange the criminals according to the order of the vowels, in the following Latin verse,

Populeam virgam mater regina ferebat.

Because o is the fourth in the order of the vowels, you must begin by four of those whom you wish to save; next to these place five of those whom you wish to punish; and so on alternately, according to the figures which stand over the vowels of the above

verse.

In a company consisting of several persons, the following game may be introduced by way of amusement.

We shall suppose that there are 13 ladies in the company; in that case, provide 12 nosegays, and in order to mortify one of them, without shewing any appearance of partiality, announce that you mean to let chance decide which of them is to go without one. For this purpose, make the 13 ladies stand up in a ring, allowing them to place themselves as they please; and distribute to them the 12 nosegays, counting them from 1 to 9, and making the ninth retire from the ring and carry with her a nosegay. It will be found, that the eleventh, reckoning from the one by whom you