Amusements in Mathematics (PDF)

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 our different digits (nought excluded) that add up to 14? Multiply 14 by the number beneath t 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 (10 n -1)) 8! divided by (9 - n)! equals the required sum. Note that 0! equals 1. This maybe 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), 4x7x6 = 168. Therefore 16,800,000 less 1,680 gives us 16,798,320 in another way.

97.?THE SPOT ON THE TABLE.? solution

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.? solution

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.? solution

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.? solution

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 p 9 161 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? solution

Add together the ten weights and divide by 4, and we get 289 lbs. as the weight of the five trusses together. If we call the five trusses in the order of weight A, B, C, D, and E, the lightest being A and the heaviest E, then the lightest, no lbs., must be the weight of A and B; and the next lightest, 112 lbs., must be the weight of A and C. Then the two heaviest, D and E, must weigh 121 lbs., and C and E must weigh 120 lbs. We thus know that A, B, D, and E weigh together 231 lbs., which, deducted from 289 lbs. (the weight of the five trusses), gives us the weight of C as 58 lbs. Now, by mere subtraction, we find the weight of each of the five trusses?54 lbs., 56 lbs., 58 lbs., 59 lbs., and 62 lbs. respectively.

102.? MR. GUBBINS IN A FOG.? solution

The candles must have burnt for three hours and three-quarters. One candle had one-sixteenth of its total length left and the other four-sixteenths.

103.?PAINTING THE LAMP-POSTS.? solution

Pat must have painted six more posts than Tim, no matter how many lamp-posts there were. For example, suppose twelve on each side; then Pat painted fifteen and Tim nine. If a hundred on each side, Pat painted one hundred and three, and Tim only ninety-seven

104.?CATCHING THE TH\EE? solution