Elements of Algebra

220 ELEMENTS SECT. IV. times 17 crowns, as the third has 5 crowns. Farther, if we multiply the money of the first by the money of the second, and the money of the second by the money of the third, and, lastly, the money of the third by that of the first, the sum of these three products will be $3830\frac{2}{3}$. How much money has each? Suppose that the first player has $x$ crowns; and since he has as many times 7 crowns, as the second has 3 crowns, we know that his money is to that of the second, in the ratio of $7:3$. We shall therefore have $7:3::x:\frac{3}{7}x$, the money of the second player. Also, as the money of the second player is to that of the third in the ratio of $17:5$, we shall have $17:5::\frac{3}{7}x:\frac{15}{119}x$, the money of the third player. Multiplying $x$, or the money of the first player, by $\frac{3}{7}x$, the money of the second, we have the product $\frac{3}{7}x^2$; then, $\frac{3}{7}x$, the money of the second, multiplied by the money of the third, or by $\frac{15}{119}x$, gives $\frac{45}{833}x^2$; and, lastly, the money of the third, or $\frac{15}{119}x$, multiplied by $x$, or the money of the first, gives $\frac{15}{119}x^2$. Now, the sum of these three products is $\frac{3}{7}x^2 + \frac{45}{833}x^2 + \frac{15}{119}x^2$; and reducing these fractions to the same denominator, we find their sum $\frac{507}{833}x^2$, which must be equal to the number $3830\frac{2}{3}$. We have therefore, $\frac{507}{833}x^2 = 3830\frac{2}{3}$. So that $\frac{1521}{833}x^2 = 11492$, and $1521x^2$ being equal to $9572836$, dividing by $1521$, we have $x^2 = \frac{9572836}{1521}$; and taking its root, we find $x = \frac{3094}{39}$. This fraction is reducible to lower terms, if we divide by $13$; so that $x = \frac{238}{3} = 79\frac{1}{3}$; and hence we conclude, that $\frac{3}{7}x = 34$, and $\frac{15}{119}x = 10$. The first player therefore has $79\frac{1}{3}$ crowns, the second has $34$ crowns, and the third $10$ crowns. Remark. This calculation may be performed in an easier manner; namely, by taking the factors of the numbers which present themselves, and attending chiefly to the squares of those factors. It is evident, that $507 = 3 \times 169$, and that $169$ is the square of $13$; then, that $833 = 7 \times 119$, and $119 = 7 \times 17$: therefore $\frac{3 \times 169}{17 \times 49}x^2 = 3830\frac{2}{3}$, and if we multiply by $3$, we have $\frac{9 \times 169}{17 \times 49}x^2 = 11492$. Let us resolve this number also into its factors; and we readily perceive, that the first is $4$; that is to say, that $11492 = 4 \times 2873$. Farther, $2873$ is divisible by $17$, so that $2873 = 17 \times 169$.