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. 116.--THE CONVERTED MISER. As we are not told in what year Mr. Jasper Bullyon made the generous distribution of his accumulated wealth, but are required to find the lowest possible amount of money, it is clear that we must look for a year of the most favourable form. There are four cases to be considered--an ordinary year with fifty-two Sundays and with fifty-three Sundays, and a leap-year with fifty-two and fifty-three Sundays respectively. Here are the lowest possible amounts in each case:-- 313 weekdays, 52 Sundays L112,055 312 weekdays, 53 Sundays 19,345 314 weekdays, 52 Sundays No solution possible. 313 weekdays, 53 Sundays L69,174 The lowest possible amount, and therefore the correct answer, is L19,345, distributed in an ordinary year that began on a Sunday. The last year of this kind was 1911. He would have paid L53 on every day of the year, or L62 on every weekday, with L1 left over, as required, in the latter event. 117.--A FENCE PROBLEM. Though this puzzle presents no great difficulty to any one possessing a knowledge of algebra, it has perhaps rather interesting features.