The method consists in filling up, as much as possible, the exterior bands, which form, as it were, a border, without entering the third, until there are no other means of passing from the square at which you have arrived to one of the two first; a rule to which the knight is necessarily subjected, in the most evident manner, from his first step to the fiftieth. When he arrives at the 50th, there is no other choice than 51 or 63; but the 51st square being nearer the border, ought to be preferred; and then his progress must necessarily be through 52, 53, 54, 55, 56, 57, 58, 59, 60, 61. When he arrives at the last, it is a matter of indifference whether he be made to pass through the three remaining squares, by directing his progress upwards or downwards; for in either case he will arrive at the last. APPLICATION OF ANALYSIS TO THE SOLUTION OF VARIOUS PROBLEMS. AS the object of this work is to unite instruction with amusement, we shall confine ourselves to such problems as are sufficiently easy to be solved by the application of those rules which we have explained in the introduction. (14.) PROBLEM I. A lady lamenting that her age was triple that of her daughter; the latter consoled her by observing, that in 15 years it would be only double: what was the age of each? Put a to denote the 15 years, and let r represent the age of the daughter; then by the conditions of the problem, the ages of the daughter and mother, which at present are r and 3x, at the end of 15 years will bera and 3 x + a; but as the age of the mother will then be double that of the daughter, we must multiply the age of the latter by 2, to have the following equation: 2 x + 2 a = 3 x + a. Then by transposition 2 a― a = 3x And then by reduction a = x. (14.) - 2x, (15.) Consequently, the age of the daughter is 15, and that of the mother 45; which will answer all the conditions of the problem. PROBLEM II. A father, on his death-bed, gave orders in his will, that if his wife, who was then pregnant, brought forth a son, he should inherit of his property, and the mother the remainder; but if she brought forth a daughter, the latter should have only, and the mother 3. As the widow, however, was delivered of twins, a boy and a girl, what share ought each to have of the property left by the father? It The only difficulty in this problem is to determine what would have been the will of the testator, had he foreseen that his wife would be delivered of twins. has generally been explained in the following manner: As the testator desired that in case his wife brought forth a son, he should have two thirds of his property, and the mother one third, it hence follows, that his intention was to give his son a sum double to that of the mother; and as he desired, in the other case, that if she brought forth a daughter, the mother should have two thirds of his property, and the daughter one third, there is reason to conclude, that he intended the share of the mother to be double that of the daughter. Consequently, to unite these two conditions, the heritage must be divided in such a manner, that the son may have twice as much as the mother, and the mother twice as much as the daughter. If a, therefore, be supposed to represent the father's property, and the share of the daughter, then 2a will express that of the mother, and 4r that of the son. But, as all these shares together are equal to the father's property, we shall have the following equation: x + 2x + 4x = a. Hence, if we suppose the whole property to be £30000. the daughter's share will be £4285; that of the mother £85713, and that of the son £171429. Sometimes the following difficulty is proposed in regard to this problem. In case the mother should be brought to bed of two sons and a daughter, in what manner must the property be divided. In our opinion, no other answer can be given than what would be given by the gentlemen of the gown, viz. that in this case the will would be void; for as no provision was made in it for a third child, its nullity would be established according to all the laws hitherto in existence. Because, 1st. The law is precise. 2nd. Because it is impossible to determine what would have been the dispositions of the testator if two sons had been born to him, or if he had foreseen that his wife would be delivered of two. PROBLEM III. A captain being asked, how many soldiers he had in his company, replied-One half of them are in camp, one third in the trenches, one eighth in the hospital, and four in prison. Of how many men did his company con sist? If the number of soldiers be expressed by r, and the four in prison by a, we shall have the following equation: Then by mult. 24x + 16x + 6x + 48a= 48x. PROBLEM IV. The head of a fish is 9 inches in length, its tail is as long as the head and half the body, and the body is as long as the head and the tail. What is the length of the fish? Let the head be expressed by a, the tail by x, and the body by y. By the conditions of the problem we shall then have the following two equations: By this result the problem is solved; for the head being supposed equal to 9, the body denoted by y= 36, and the tail, being equal to the head and half the body, must necessarily be 27, which answers all the conditions of the problem. PROBLEM V. A person who had a lease of a house for 99 years, being asked when it would expire, replied, that two-thirds of the time he had possessed it were exactly equal to fourfifths of the time unexpired. How many years of the lease were still remaining? If we call the time elapsed x, and the 99 years a, the time unexpired will be a x. Therefore, by the conditions of the problem, ora 54. |