Part 69

The distance must be 63/4 miles.

71.--SIR EDWYN DE TUDOR.

The distance must have been sixty miles. If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock--an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock--an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock--the time appointed.

72.--THE HYDROPLANE QUESTION.

The machine must have gone at the rate of seven-twenty-fourths of a mile per minute and the wind travelled five-twenty-fourths of a mile per minute. Thus, going, the wind would help, and the machine would do twelve-twenty-fourths, or half a mile a minute, and returning only two-twenty-fourths, or one-twelfth of a mile per minute, the wind being against it. The machine without any wind could therefore do the ten miles in thirty-four and two-sevenths minutes, since it could do seven miles in twenty-four minutes.

73.--DONKEY RIDING.

The complete mile was run in nine minutes. From the facts stated we cannot determine the time taken over the first and second quarter-miles separately, but together they, of course, took four and a half minutes. The last two quarters were run in two and a quarter minutes each.

74.--THE BASKET OF POTATOES.

Multiply together the number of potatoes, the number less one, and twice the number less one, then divide by 3. Thus 50, 49, and 99 multiplied together make 242,550, which, divided by 3, gives us 80,850 yards as the correct answer. The boy would thus have to travel 45 miles and fifteen-sixteenths--a nice little recreation after a day's work.

75.--THE PASSENGER'S FARE.

Mr. Tompkins should have paid fifteen shillings as his correct share of the motor-car fare. He only shared half the distance travelled for L3, and therefore should pay half of thirty shillings, or fifteen shillings.

76.--THE BARREL OF BEER.

Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which together sum to 29, whose digital root is 2. As the contents of the barrels sold must be a number divisible by 3, if one buyer purchased twice as much as the other, we must find a barrel with root 2, 5, or 8 to set on one side. There is only one barrel, that containing 20 gallons, that fulfils these conditions. So the man must have kept these 20 gallons of beer for his own use and sold one man 33 gallons (the 18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the 16, 19, and 31 gallon barrels).

77.--DIGITS AND SQUARES.

The top row must be one of the four following numbers: 192, 219, 273, 327. The first was the example given.

78.--ODD AND EVEN DIGITS.

As we have to exclude complex and improper fractions and recurring decimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, both equal 84+1/3. Without any use of fractions it is obviously impossible.

79.--THE LOCKERS PUZZLE.

The smallest possible total is 356 = 107 + 249, and the largest sum possible is 981 = 235 + 746, or 657+324. The middle sum may be either 720 = 134 + 586, or 702 = 134 + 568, or 407 = 138 + 269. The total in this case must be made up of three of the figures 0, 2, 4, 7, but no sum other than the three given can possibly be obtained. We have therefore no choice in the case of the first locker, an alternative in the case of the third, and any one of three arrangements in the case of the middle locker. Here is one solution:--

107 134 235 249 586 746 --- --- --- 356 720 981

Of course, in each case figures in the first two lines may be exchanged vertically without altering the total, and as a result there are just 3,072 different ways in which the figures might be actually placed on the locker doors. I must content myself with showing one little principle involved in this puzzle. The sum of the digits in the total is always governed by the digit omitted. 9/9 - 7/10 - 5/11 -3/12 - 1/13 - 8/14 - 6/15 - 4/16 - 2/17 - 0/18. Whichever digit shown here in the upper line we omit, the sum of the digits in the total will be found beneath it. Thus in the case of locker A we omitted 8, and the figures in the total sum up to 14. If, therefore, we wanted to get 356, we may know at once to a certainty that it can only be obtained (if at all) by dropping the 8.

80.--THE THREE GROUPS.

There are nine solutions to this puzzle, as follows, and no more:--

12 x 483 = 5,796 27 x 198 = 5,346 42 x 138 = 5,796 39 x 186 = 7,254 18 x 297 = 5,346 48 x 159 = 7,632 28 x 157 = 4,396 4 x 1,738 = 6,952 4 x 1,963 = 7,852

The seventh answer is the one that is most likely to be overlooked by solvers of the puzzle.

81.--THE NINE COUNTERS.

In this case a certain amount of mere "trial" is unavoidable. But there are two kinds of "trials"--those that are purely haphazard, and those that are methodical. The true puzzle lover is never satisfied with mere haphazard trials. The reader will find that by just reversing the figures in 23 and 46 (making the multipliers 32 and 64) both products will be 5,056. This is an improvement, but it is not the correct answer. We can get as large a product as 5,568 if we multiply 174 by 32 and 96 by 58, but this solution is not to be found without the exercise of some judgment and patience.

82.--THE TEN COUNTERS.