Elements of Algebra

CHAP. I. OF ALGEBRA. 307 and, lastly, $N = 39 \times 56u + 9883$*. And the least possible value of N is found by making $u = -4$; for by this supposition, we have $N = 1147$: and if we make $u = x-4$, we find $N = 2184x - 8736 + 9883$; or $N = 2184x + 1147$; which numbers form an arithmetical progression, whose first term is 1147, and whose common difference is 2184; the following being some of its leading terms: 1147, 3331, 5515, 7699, 9883, &c. 15. We shall subjoin some other questions by way of practice. *Question* 9. A company of men and women club together for the payment of a reckoning: each man pays 25 livres, and each woman 16 livres; and it is found that all the women together have paid 1 livre more than the men. How many men and women were there? Let the number of women be $p$, and that of men $q$; then the women will have expended $16p$, and the men $25q$; so that $16p = 25q + 1$, and $p = \frac{25q+1}{16} = q + \frac{9q+1}{16} = q + r$, or $16r = 9q + 1$, $q = \frac{16r-1}{9} = r + \frac{7r-1}{9} = r + s$, or $9s = 7r - 1$, $r = \frac{9s+1}{7} = s + \frac{2s+1}{7} = s + t$, or $7t = 2s + 1$, $s = \frac{7t-1}{2} = 3t + \frac{t-1}{2} = 3t + u$, or $2u = t - 1$. We shall therefore obtain, by tracing back our substitutions, $t = 2u + 1$, $s = 3t + u = 7u + 3$, $r = s + t = 9u + 4$, $q = r + s = 16u + 7$, $p = q + r = 25u + 11$. So that the number of women was $25u + 11$, and that of men was $16u + 7$; and in these formulæ we may substitute * As the numbers 176 and 253 ought, respectively, to be divisible by 39 and 56; and as the former ought, by the question, to leave the remainder 16, and the latter 27, the sum 9883 is formed by multiplying 176 by 56, and adding the remainder 27 to the product: or by multiplying 253 by 39, and adding the remainder 16 to the product. Thus, $(176 \times 56) + 27 = 9883$; and $(253 \times 39) + 16 = 9883$. x 2