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.

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.