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.