Suppose he fixes upon five numbers, viz. 4, 6, 9, 15, 16.
He must add together the numbers as follows, and tell you the various sums:
| 1. | The sum of the 1st and 2d | 10 |
| 2. | The sum of the 2d and 3d | 15 |
| 3. | The sum of the 3d and 4th | 24 |
| 4. | The sum of the 4th and 5th | 31 |
| 5. | The sum of the 1st and last | 20 |
You must then add together the 1st, 3d and 5th sums, viz. 10 + 24 + 20 = 54, and the 2d and 4th, 15 + 31 = 46; take one from the other, leaving 8. The half of this is the 1st number, 4; if you take this from the sum of the 1st and 2d you will have the 2d number, 6; this taken from the sum of the 2d and 3d will give you the 3d, 9; and so on for the other numbers.
3d Case.—Where one or more of the numbers are 10, or more than 10, and where an even number of numbers has been thought of.
Suppose he fixes on six numbers, viz. 2, 6, 7, 15, 16, 18. He must add together the numbers as follows, and tell you the sum in each case:—
| 1. | The sum of the 1st and 2d | 8 |
| 2. | The sum of the 2d and 3d | 13 |
| 3. | The sum of the 3d and 4th | 22 |
| 4. | The sum of the 4th and 5th | 31 |
| 5. | The sum of the 5th and 6th | 34 |
| 6. | The sum of the 2d and last | 24 |
You must then add together the 2d, 4th and 6th sums, 13 + 31 + 24 = 68, and the 3d and 5th sums, 22 + 34 = 56. Subtract one from the other, leaving 12; the 2d number will be 6, the half of this; take the 2d from the sum of the 1st and 2d you will get the 1st; take the 2nd from the sum of the 2d and 3d, and you will have the 3d, and so on.
HOW MANY COUNTERS HAVE I IN MY HANDS?
A person having an equal number of counters in each hand, it is required to find how many he has altogether.