111.—­REAPING THE CORN.

The whole field must have contained 46.626 square rods.  The side of the central square, left by the farmer, is 4.8284 rods, so it contains 23.313 square rods.  The area of the field was thus something more than a quarter of an acre and less than one-third; to be more precise, .2914 of an acre.

112.—­A PUZZLING LEGACY.

As the share of Charles falls in through his death, we have merely to divide the whole hundred acres between Alfred and Benjamin in the proportion of one-third to one-fourth—­that is in the proportion of four-twelfths to three-twelfths, which is the same as four to three.  Therefore Alfred takes four-sevenths of the hundred acres and Benjamin three-sevenths.

113.—­THE TORN NUMBER.

The other number that answers all the requirements of the puzzle is 9,801.  If we divide this in the middle into two numbers and add them together we get 99, which, multiplied by itself, produces 9,801.  It is true that 2,025 may be treated in the same way, only this number is excluded by the condition which requires that no two figures should be alike.

The general solution is curious.  Call the number of figures in each half of the torn label n.  Then, if we add 1 to each of the exponents of the prime factors (other than 3) of 10^n — 1 (1 being regarded as a factor with the constant exponent, 1), their product will be the number of solutions.  Thus, for a label of six figures, n = 3.  The factors of 10^n - 1 are 1¹ x 37¹ (not considering the 3 cubed), and the product of 2 x 2 = 4, the number of solutions.  This always includes the special cases 98 — 01, 00 — 01, 998 — 01, 000 — 001, etc.  The solutions are obtained as follows:—­Factorize 10 cubed — 1 in all possible ways, always keeping the powers of 3 together, thus, 37 x 27, 999 x 1.  Then solve the equation 37x = 27y + 1.  Here x = 19 and y = 26.  Therefore, 19 x 37 = 703, the square of which gives one label, 494,209.  A complementary solution (through 27x = 37x + 1) can at once be found by 10^n — 703 = 297, the square of which gives 088,209 for second label. (These non-significant noughts to the left must be included, though they lead to peculiar cases like 00238 — 04641 = 4879 squared, where 0238 — 4641 would not work.) The special case 999 x 1 we can write at once 998,001, according to the law shown above, by adding nines on one half and noughts on the other, and its complementary will be 1 preceded by five noughts, or 000001.  Thus we get the squares of 999 and 1.  These are the four solutions.

114.—­CURIOUS NUMBERS.

The three smallest numbers, in addition to 48, are 1,680, 57,120, and 1,940,448.  It will be found that 1,681 and 841, 57,121 and 28,561, 1,940,449 and 970,225, are respectively the squares of 41 and 29, 239 and 169, 1,393 and 985.

115.—­A PRINTER’S ERROR.

The answer is that 2^5 .9^2 is the same as 2592, and this is the only possible solution to the puzzle.