Elements of Algebra

310 ELEMENTS PART II. may immediately determine $p$ and $q$ by the succeeding letters, as follows: Let $\left\{ \begin{array}{l} a = Ab + c \\ b = Bc + d \\ c = Cd + e \\ d = De + f \\ e = Ef + g \\ f = Fg + 0 \end{array} \right.$ and we shall find $\left\{ \begin{array}{l} p = Aq + r \\ q = Br + s \\ r = Cs + t \\ s = Dt + u \\ t = Eu + v \\ u = Fv \pm n. \end{array} \right.$ We have only to observe farther, that in the last equation, the sign $+$ must be prefixed to $n$, when the number of equations is odd; and that, on the contrary, we must take $-n$, when the number is even: by these means, the questions which form the subject of the present chapter may be readily answered, of which we shall give some examples. 20. *Question* 11. Required a number, which, being divided by 11, leaves the remainder 3; but being divided by 19, leaves the remainder 5. Call this number N; then, in the first place, we have $N = 11p + 3$, and in the second, $N = 19q + 5$; therefore, we have the equation $11p = 19q + 2$, which furnishes the following Table: $19 = 1 \times 11 + 8$ | $p = q + r$ $11 = 1 \times 8 + 3$ | $q = r + s$ $8 = 2 \times 3 + 2$ | $r = 2s + t$ $3 = 1 \times 2 + 1$ | $s = t + u$ $2 = 2 \times 1 + 0$ | $t = 2u$ where we may assign any value to $u$, and determine by it the preceding letters successively. We find, $t \ldots = 2u + 2$ $s = t + u = 3u + 2$ $r = 2s + t = 8u + 6$ $q = r + s = 11u + 8$ $p = q + r = 19u + 14$; whence, taking $u = 0$, we obtain the number sought $N = 11p + 3 = 11(19u + 14) + 3 = 209u + 157$; therefore 157 is the least number that can express N, or satisfy the terms of the question *. 21. *Question* 12. To find a number N such, that if we divide it by 11, there remains 3, and if we divide it by 19, there remains 5; and farther, if we divide it by 29, there remains 10. The last condition requires that $N = 29p + 10$; and as we have already performed the calculation (in the last * Because, in this case, $u = 0$.