Method.—The form of the cross when left is represented by Fig. 1, and when returned by Fig. 2. It will be seen by the figures how the diamonds were counted by the old Jew, and how they were arranged by the jeweler, who “jewed” the Jew.
Take 10 pieces of money, lay them in a row, and require some one to put them together into heaps of two in each heap by passing each piece over two others.
Method.—Let the pieces be denoted by the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Then place 7 on 10, 5 on 2, 3 on 8, 1 on 4, and 9 on 6.
A man goes to a store and purchases a pair of boots worth $5 and hands out a $50 bill to pay for them. The merchant, not being able to make the change, goes over the street to a broker and gets the bill changed and then returns and gives the man who bought the boots his change. After the purchaser of the boots had been gone a few hours the broker, finding the bill to be a counterfeit, comes and demands $50 of good money from the merchant. How much does the merchant lose?
Remark.—At first glance some say $45 and the boots, some $50 and the boots, some $95 and the boots, and others $100 and the boots. Which is correct?
A vessel with a crew of 30 men, half of whom were black, became short of provisions and fearing that unless half the crew were thrown overboard all would perish, the captain proposed to the sailors to stand upon deck in a row and every ninth man be thrown overboard until half the crew were destroyed. It so happened that the whites were saved. Required, the order of arrangement.
Answer.—W W W W B B B B B W W B W W W B W B B W W B B B W B B W W B. This can easily be proved by trial, using letters or figures to represent men.
Suppose a hare is 10 rods before a hound, and that the hound runs 10 rods while the hare runs 1 rod. Now, when the hound has run 10 rods the hare has run 1 rod; hence they are now 1 rod apart, and when the hound has run that one rod the hare has run 1/10 of a rod; hence they are now 1/10 of a rod apart, and when the hound has run the 1/10 of a rod they are 1/100 of a rod apart; and in the same way it may be shown the hare is always 1/10 of the previous distance ahead of the hound; hence the hound can never catch the hare. How is the contrary shown mathematically? How far will the hound run to catch the hare.
Answer.—The distance the hound runs will be represented by the series
| 1 | 1 | 1 | |||
| 10 + 1 + | —— | + | —— | + | ——, |
| 10 | 100 | 1000 |