Elements of Algebra

308 ELEMENTS PART II. for $u$ any integer numbers whatever. The least results, therefore, will be as follow: Number of women, $11, 36, 61, 86, 111, \ldots$. of men, $7, 23, 39, 55, 71, \ldots$. According to the first answer, or that which contains the least numbers, the women expended 176 livres, and the men 175 livres; that is, one livre less than the women. 16. Question 10. A person buys some horses and oxen: he pays 31 crowns per horse, and 20 crowns for each ox; and he finds that the oxen cost him 7 crowns more than the horses. How many oxen and horses did he buy? If we suppose $p$ to be the number of the oxen, and $q$ the number of the horses, we shall have the following equation: $p = \frac{31q+7}{20} = q + \frac{11q+7}{20} = q + r$, or $20r = 11q+7$, $q = \frac{20r-7}{11} = r + \frac{9r-7}{11} = r+s$, or $11s = 9r-7$, $r = \frac{11s+7}{9} = s + \frac{2s+7}{9} = s+t$, or $9t = 2s+7$, $s = \frac{9t-7}{2} = 4t + \frac{t-7}{2} = 4t+u$, or $2u = t-7$, whence $t \ldots = 2u+7$, and, consequently, $s = 4t+u = 9u+28$, $r = s+t = 11u+35$, $q = r+s = 20u+63$, number of horses, $p = q+r = 31u+98$, number of oxen. Whence, the least positive values of $p$ and $q$ are found by making $u = -3$; those which are greater succeed in the following arithmetical progressions: Number of oxen, $p = 5, 36, 67, 98, 129, 160, 191, 222, 253, \ldots$. Number of horses, $q = 3, 23, 43, 63, 83, 103, 123, 143, 163, \ldots$. 17. If now we consider how the letters $p$ and $q$, in this example, are determined by the succeeding letters, we shall perceive that this determination depends on the ratio of the numbers 31 and 20, and particularly on the ratio which we discover by seeking the greatest common divisor of these two numbers. In fact, if we perform this operation,