A puzzling Question to be proposed for Solution.

Set down three sums on paper; and say to the company, ladies and gentlemen, there are three sums, very different from each other, and very disproportionate; yet I wish to divide them among three persons, so that they may have an equal sum each, and yet without altering any thing in either of the sums. This will appear very difficult, yet nothing so simple and easy; one single addition will suffice to prove to you that the amount of each sum will be the same, and that the shares will not enrich much the respective persons: here is the proof:

EXAMPLE.

OPERATION.

Cast up the first of these sums in the following manner, and say, 5 and 1 make 6; 3 more, 9; 4 more, 13; 1 more, 14; 2 more, 16; and 2 more, 18: set down——18.

Make the addition of the second sum in the same manner as you have done the first, and you will find the same sum of——18.

Then proceed for the third as in the two preceding, and the product will be also——18.

Here then is my division made, and each person will have only 18, as I have proved by the foregoing example.

By this we see, that nothing more is required than to be attentive in setting the sums, to make the numbers so that each sum may amount only to 18.

You may make the same question on whatever sum you please, only observing, as above, that the amount of the numbers you set may not exceed the sum you desire to belong to each person that is to have a share.