Elements of Algebra

CHAP. I. OF ALGEBRA. 305 ner, $35z = 9u$, and $u = \frac{35z}{9}$; where it is evident that $z$ must be divisible by 9; therefore let $z = 9s$; and we shall then have $u = 35s$, and N the number sought $= 315s$. 12. We find more difficulty, when $c$ is not $= 0$. For example, when $5x = 7y + 3$, the equation to which we are led, and which requires us to seek a number N such, that it may be divisible by 5, and if divided by 7, may leave the remainder 3: for we must then have $N = 5x$, and also $N = 7y + 3$, whence results the equation $5x = 7y + 3$; and, consequently, $x = \frac{7y+3}{5} = \frac{5y+2y+3}{5} = y + \frac{2y+3}{5}$ If we make $2y + 3 = 5z$, or $z = \frac{2y+3}{5}$, we have $x = y + z$. Now, because $2y + 3 = 5z$, or $2y = 5z - 3$, we have $y = \frac{5z-3}{2}$, or $y = 2z + \frac{z-3}{2}$. If, therefore, we farther suppose $z - 3 = 2u$, we have $z = 2u + 3$, and $y = 5u + 6$, and $x = y + z = (5u + 6) + (2u + 3) = 7u + 9$. Hence, the number sought $N = 35u + 45$, in which equation we may substitute for $u$ not only all positive integer numbers, but also negative numbers; for, as it is sufficient that N be positive, we may make $u = -1$, which gives $N = 10$; the other values are obtained by continually adding 35; that is to say, the numbers sought are 10, 45, 80, 115, 150, 185, 220, \ldots. 13. The solution of questions of this sort depends on the relation of the two numbers by which we are to divide; that is, they become more or less tedious, according to the nature of those divisors. The following question, for example, admits of a very short solution: Required a number which, divided by 6, leaves the remainder 2; and divided by 13, leaves the remainder 3. Let this number be N. First, $N = 6x + 2$, and then $N = 13y + 3$; consequently, $6x + 2 = 13y + 3$, and $6x = 13y + 1$; hence, $x = \frac{13y+1}{6} = 2y + \frac{y+1}{6}$, and if we make $y + 1 = 6z$, or $y = 6z - 1$, we obtain $x = 2y + z = 13z - 2$; whence we have for the number x