Page:Amusements in mathematics.djvu/174

162 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.

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$3$ are 1$1$×37$1$ (not considering the 3$3$), and the product of 2×2=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$3$—1 in all possible ways, always keeping the powers of 3 together, thus, 37×27, 999×1. Then solve the equation $$37x=27y+1$$. Here $$x$$=19 and $$y$$=26. Therefore, 19×37=703, the square of which gives one label, 494, 209. A complementary solution (through $$27x=37y+1$$) 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 = 48792, where 0238—4641 would not work.) The special case 999×1 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.

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.

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.

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 fiftytwo and fifty-three Sundays respectively. Here are the lowest possible amounts in each case:—

The lowest possible amount, and therefore 1 the correct answer, is £19,345, distributed in an ordinary year that began on a Sunday. The last year of this kind was 1911. He would have paid £53 on every day of the year, or £62 on every weekday, with £1 left over, as required, in the latter event.

117.— A FENCE PROBLEM.

this puzzle presents no great difficulty to any one possessing a knowledge of algebra, it has perhaps rather interesting features.

Seeing, as one does in the illustration, just one corner of the proposed square, one is scarcely prepared for the fact that the field, in order to comply with the conditions, must contain exactly 501,760 acres, the fence requiring the same number of rails. Yet this is the correct answer, and the only answer, and if that gentleman in Iowa carries out his intention, his field will be twenty-eight miles long on each side, and a little larger than the county of Westmorland. I am not aware that any limit has ever been fixed to the size of a "field," though they do not run so large as this in Great Britain. Still, out in Iowa, where my correspondent resides, they do these things on a very big scale. I have, however, reason to believe that when he finds the sort of task he has set himself, he will decide to abandon it; for if that cow decides to roam to fresh woods and pastures new, the milkmaid may have to start out a week in advance in order to obtain the morning's milk.

Here is a little rule that will always apply where the length of the rail is half a pole. Multiply the number of rails in a hurdle by four, and the result is the exact number of miles in the side of a square field containing the same number of acres as there are rails in the complete fence. Thus, with a one-rail fence the field is four miles square; a two-rail fence gives eight miles square; a three-rail fence, twelve miles square; and so on, until we find that a seven-rail fence multiplied by four gives a field of twenty-eight miles square. In the case of our present problem, if the field be made smaller, then the number of rails will exceed the number of acres; while if the field be made larger, the number of rails will be less than the acres of the field.

118.—CIRCLING THE SQUARES.

this problem might strike the novice as being rather difficult, it is, as a matter of fact, quite easy, and is made still easier by inserting four out of the ten numbers.