| Signs. | Strokes. | ||
| 1+(2×3)+4+5+67+8+9=100 | (8 | .. | 16) |
| (1×2)+34+56+7-8+9=100 | (7 | .. | 13) |
| 12+3-4+5+67+8+9=100 | (6 | .. | 11) |
| 123-4-5-6-7+8-9=100 | (6 | .. | 7) |
| 123+4-5+67-8-9=100 | (4 | .. | 6) |
| 123+45-67+8-9=100 | (4 | .. | 6) |
| 123-45-67+89=100 | (3 | .. | 4) |
It will be noticed that in the above I have counted the bracket as one sign and two strokes. The last solution is singularly simple, and I do not think it will ever be beaten.
95.—THE FOUR SEVENS.
The way to write four sevens with simple arithmetical signs so that they represent 100 is as follows:—
7.7 × 7.7 = 100.
Of course the fraction, 7 over decimal 7, equals 7 divided by 710, which is the same as 70 divided by 7, or 10. Then 10 multiplied by 10 is 100, and there you are! It will be seen that this solution applies equally to any number whatever that you may substitute for 7.
96.—THE DICE NUMBERS.
The sum of all the numbers that can be formed with any given set of four different figures is always 6,666 multiplied by the sum of the four figures. Thus, 1, 2, 3, 4 add up 10, and ten times 6,666 is 66,660. Now, there are thirty-five different ways of selecting four figures from the seven on the dice—remembering the 6 and 9 trick. The figures of all these thirty-five groups add up to 600. Therefore 6,666 multiplied by 600 gives us 3,999,600 as the correct answer.
Let us discard the dice and deal with the problem generally, using the nine digits, but excluding nought. Now, if you were given simply the sum of the digits—that is, if the condition were that you could use any four figures so long as they summed to a given amount—then we have to remember that several combinations of four digits will, in many cases, make the same sum.
| 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| 1 | 1 | 2 | 3 | 5 | 6 | 8 | 9 | 11 | 11 | 12 |
| 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | |
| 11 | 11 | 9 | 8 | 6 | 5 | 3 | 2 | 1 | 1 |
Here the top row of numbers gives all the possible sums of four different figures, and the bottom row the number of different ways in which each sum may be made. For example, 13 may be made in three ways: 1237, 1246, and 1345. It will be found that the numbers in the bottom row add up to 126, which is the number of combinations of nine figures taken four at a time. From this table we may at once calculate the answer to such a question as this: What is the sum of all the numbers composed of four different digits (nought excluded) that add up to 14? Multiply 14 by the number beneath it in the table, 5, and multiply the result by 6,666, and you will have the answer. It follows that, to know the sum of all the numbers composed of four different digits, if you multiply all the pairs in the two rows and then add the results together, you will get 2,520, which, multiplied by 6,666, gives the answer 16,798,320.
The following general solution for any number of digits will doubtless interest readers. Let n represent number of digits, then 5 (10n—1) 8 divided by 9-n equals the required sum. Note that 0 equals 1. This may be reduced to the following practical rule: Multiply together 4×7×6×5... to (n-1) factors; now add (n+1) ciphers to the right, and from this result subtract the same set of figures with a single cipher to the right. Thus for n=4 (as in the case last mentioned), 4×7×6=168. Therefore 16,800,000 less 1,680 gives us 16,798,320 in another way.
97.—THE SPOT ON THE TABLE.
The ordinary schoolboy would correctly treat this as a quadratic equation. Here is the actual arithmetic. Double the product of the two distances from the walls. This gives us 144, which is the square of 12. The sum of the two distances is 17. If we add these two numbers, 12 and 17, together, and also subtract one from the other, we get the two answers that 29 or 5 was the radius, or half-diameter, of the table. Consequently, the full diameter was 58 in. or 10 in. But a table of the latter dimensions would be absurd, and not at all in accordance with the illustration. Therefore the table must have been 58 in. in diameter. In this case the spot was on the edge nearest to the comer of the room—to which the boy was pointing. If the other answer were admissible, the spot would be on the edge farthest from the comer of the room.
98.—ACADEMIC COURTESIES.
There must have been ten boys and twenty girls. The number of bows girl to girl was therefore 380, of boy to boy 90, of girl with boy 400, and of boys and girls to teacher 30, making together 900, as stated. It will be remembered that it was not said that the teacher himself returned the bows of any child.
99.—THE THIRTY-THREE PEARLS.
The value of the large central pearl must have been £3,000. The pearl at one end (from which they increased in value by £100) was £1,400; the pearl at the other end, £600.
100.—THE LABOURER'S PUZZLE.
The man said, "I am going twice as deep," not "as deep again." That is to say, he was still going twice as deep as he had gone already, so
