The following general solution for any number of digits will doubtless interest readers. Let n represent number of digits, then 5 (10^n - 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 x 7 x 6 x 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 x 7 x 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 corner 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 corner 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 L3,000. The pearl at one end (from which they increased in value by L100) was L1,400; the pearl at the other end, L600. 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 that when finished the hole would be three times its present depth. Then the answer is that at present the hole is 3 ft. 6 in. deep and the man 2 ft. 4 in. above ground. When completed the hole will be 10 ft. 6 in. deep, and therefore the man will then be 4 ft. 8 in. below the surface, or twice the distance that he is now above ground. 101.--THE TRUSSES OF HAY.