began, will remain the last; and consequently will have no share in the distribution.

The following table will shew the person, before her whom you wish to exclude, with whom you must begin to count 9, supposing always that the number of the nosegays is less by 1 than that of the persons.

1 For 13 persons, the 11th before.

2d.

5th.

7th.

A man has a wolf, a goat, and a cabbage, to carry over a river, but, as he is obliged to transport them one by one, in what manner is this to be done, that the wolf may not be left with the goat, nor the goat with the cabbage?

He must first carry over the goat, and then return for the wolf; when he carries over the wolf, he must take back with him the goat, which he must leave, in order to carry over the cabbage; he may then return, and carry over the goat. By these means the wolf will never be left with the goat, nor the goat with the cabbage, but when the boatman is present.

PROBLEM XXV.

In what manner can counters be disposed in the eight external cells of a square, so that there may always be 9 in each row, and yet the whole number shall vary from 20 to 32?

This problem may be proposed in the following manner:-A wine merchant caused 32 casks of choice

wine to be deposited in his cellar, giving orders to his clerk to arrange them in the annexed figure, so that each external row should contain 9.

The clerk, however, took away 12 of them, at three different times; that is, 4 each time; yet when the merchant went into the cellar, after each theft had been committed, the clerk always made him count 9 in each row. How was this possible?

This problem may be easily solved by inspecting the following figures:

PROBLEM XXVI.

To distribute among 3 persons, 21 casks of wine, 7 of them full, 7 of them empty, and 7 of them half full, so that each of them shall have the same quantity of wine, and the same number of casks.

This problem admits of two solutions, which may be clearly comprehended by means of the two following

A schoolmaster, to amuse his scholars, shewed them a number, which he said was the sum of 6 rows, each consisting of 4 figures; he then desired them to write down 3 rows of figures, to which he would add 3 more, and assured them that the sum of the whole should be equal to the number he shewed them.

To solve this problem, multiply in your own mind 9999 by 3; and the product 29997 will be the number which the schoolmaster shewed to his scholars.

It may be here seen, that each figure set down by the master is the complement to 9 of that set down by the scholars; and consequently the sum, though written down beforehand, must be exact.

PROBLEM XXVIII.

Having desired any person to multiply, for example, one of the three following numbers, by any figure at pleasure, and to tell you the product, after suppressing one figure of it, and even changing the order of the rest, to guess the figure that has been suppressed.

Let the three given numbers be

364851

234765

823644

If we suppose the person to multiply the third number by 6, the product of which will be 4941864, whatever figure be effaced, it may be easily discovered by that wanting to complete the product, as the sum of its figures must necessarily be a multiple of 9. If the 6, for example, be suppressed, the sum will not be a multiple of 9; for it amounts only to 30: as 6 therefore is wanting to 30 to make it a multiple of 9, you may boldly assert that 6 has been suppressed.

As the sum of the figures would still be a multiple of 9, if a cypher were suppressed, and as it would consequently have no need of being complete, you must make it a condition of the problem that the person shall suppress only one significant or effective figure; and if you find that the sum has no need of being completed, you may conclude that the figure suppressed has been a 9.

A mountebank, to give the greater air of the marvellous to this sport, pretended to discover by the smell what figure had been suppressed; but it may easily be supposed, that while he pretended to smell the figures, he privately added them together, so as to discover their sum.

There is another method of guessing the suppressedfigure, even when the person has been allowed to write down the sum to be multiplied himself; but in this case you must stipulate to have permission to add any one figure you choose: you must observe what figure is wanting to complete the sum, and set down that figure; if nothing is wanting, you may add 0, or 9.

PROBLEM XXIX.

A person having made choice of two numbers, and multiplied them together, to tell the product, provided you know only the last figure of it.

Have in readiness a small bag with two divisions, and put into one of them 12 square bits of card, each inscribed with the number 73; and into the second 9 other pieces, inscribed with the terms of the arithmetical progression, 3, 6, 9, 12, 15, 18, 21, 24, 27.

Present that aperture of the bag which contains the numbers 73, and desire the person to draw out one; then dexterously change the side of the bag, and having desired another person to draw any number from the second division, bid him multiply the number he has taken by that drawn out by the first person: the product will necessarily be one of the nine numbers 219, 438, 657, 876, 1095, 1314, 1533, 1752, 1971. You may then easily tell the product of the multiplication, if you know only the last figure of it.

It must here be observed, that this recreation requires a good memory; as it will be necessary to know by heart the above nine products. The following, founded on the same principle, is much easier.

PROBLEM XXX.

A person having chosen two numbers, and divided the greater by the less, to tell the quotient; that is to say, how many times the less is contained in the greater.

Put into the first division of the bag the nine numbers, 219, 438, 657, 876, 1095 1314, 1533, 1752,