§ 3. The Sons' Ages.
Problem.—"At first, two of the ages are together equal to the third. A few years afterwards, two of them are together double of the third. When the number of years since the first occasion is two-thirds of the sum of the ages on that occasion, one age is 21. What are the other two?
Answer.—"15 and 18."
Solution.—Let the ages at first be x, y, (x + y). Now, if a + b = 2c, then (a - n) + (b - n) = 2(c - n), whatever be the value of n. Hence the second relationship, if ever true, was always true. Hence it was true at first. But it cannot be true that x and y are together double of (x + y). Hence it must be true of (x + y), together with x or y; and it does not matter which we take. We assume, then, (x + y) + x = 2y; i.e. y = 2x. Hence the three ages were, at first, x, 2x, 3x; and the number of years, since that time is two-thirds of 6x, i.e. is 4x. Hence the present ages are 5x, 6x, 7x. The ages are clearly integers, since this is only "the year when one of my sons comes of age." Hence 7x = 21, x = 3. and the other ages are 15, 18.
