in four sties so that, as you go round and round, you may always find the number in each sty nearer to ten than the number in the last.
Answer.—Place 8 pigs in the first sty, 10 in the second, nothing in the third, and 6 in the fourth: 10 is nearer ten than 8; nothing is nearer ten than 10; 6 is nearer ten than nothing; and 8 is nearer ten than 6.
This problem is noticed by only two correspondents. Balbus says "it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble." Nolens Volens makes Her Radiancy change the direction of going round; and even then is obliged to add "the pigs must be carried in front of her"!
§ 2. The Grurmstipths.
Problem.—Omnibuses start from a certain point, both ways, every 15 minutes. A traveller, starting on foot along with one of them, meets one in 12½ minutes: when will he be overtaken by one?
Answer.—In 6¼ minutes.
Solution. —Let " a " be the distance an omnibus goes in 15 minutes, and " x " the distance from the starting-point to where the traveller is overtaken. Since the omnibus met is due at the starting-point in 2½ minutes, it goes in that time as far as the traveller walks in 12½; i.e. it goes 5 times as fast. Now the overtaking omnibus is " a " behind the traveller when he starts, and therefore goes " a + x " while he goes " x ." Hence a + x = 5 x ; i.e. 4 x = a , and x = a /4. This distance would be traversed