122.—­THE SEE-SAW PUZZLE.

The boy’s weight must have been about 39.79 lbs.  A brick weighed 3 lbs.  Therefore 16 bricks weighed 48 lbs. and 11 bricks 33 lbs.  Multiply 48 by 33 and take the square root.

123.—­A LEGAL DIFFICULTY.

It was clearly the intention of the deceased to give the son twice as much as the mother, or the daughter half as much as the mother.  Therefore the most equitable division would be that the mother should take two-sevenths, the son four-sevenths, and the daughter one-seventh.

124.—­A QUESTION OF DEFINITION.

There is, of course, no difference in area between a mile square and a square mile.  But there may be considerable difference in shape.  A mile square can be no other shape than square; the expression describes a surface of a certain specific size and shape.  A square mile may be of any shape; the expression names a unit of area, but does not prescribe any particular shape.

125.—­THE MINERS’ HOLIDAY.

Bill Harris must have spent thirteen shillings and sixpence, which would be three shillings more than the average for the seven men—­half a guinea.

126.—­SIMPLE MULTIPLICATION.

The number required is 3,529,411,764,705,882, which may be multiplied by 3 and divided by 2, by the simple expedient of removing the 3 from one end of the row to the other.  If you want a longer number, you can increase this one to any extent by repeating the sixteen figures in the same order.

127.—­SIMPLE DIVISION.

Subtract every number in turn from every other number, and we get 358 (twice), 716, 1,611, 1,253, and 895.  Now, we see at a glance that, as 358 equals 2 x 179, the only number that can divide in every case without a remainder will be 179.  On trial we find that this is such a divisor.  Therefore, 179 is the divisor we want, which always leaves a remainder 164 in the case of the original numbers given.

128.—­A PROBLEM IN SQUARES.

The sides of the three boards measure 31 in., 41 in., and 49 in.  The common difference of area is exactly five square feet.  Three numbers whose squares are in A.P., with a common difference of 7, are 113/120, 337/120, 463/120; and with a common difference of 13 are 80929/19380, 106921/19380, and 127729/19380.  In the case of whole square numbers the common difference will always be divisible by 24, so it is obvious that our squares must be fractional.  Readers should now try to solve the case where the common difference is 23.  It is rather a hard nut.

129.—­THE BATTLE OF HASTINGS.

Any number (not itself a square number) may be multiplied by a square that will give a product 1 less than another square.  The given number must not itself be a square, because a square multiplied by a square produces a square, and no square plus 1 can be a square.  My remarks throughout must be understood to apply to whole numbers, because fractional soldiers are not of much use in war.