Elements of Algebra

300 ELEMENTS PART II. times, there may be an infinite number, but such as are not readily obtained; and sometimes, also, none of them are possible. Hence it happens, that this part of analysis frequently requires artifices entirely appropriate to it, which are of great service in exercising the judgment of beginners, and giving them dexterity in calculation. 3. To begin with one of the easiest questions. Let it be required to find two positive, integer numbers, the sum of which shall be equal to 10. Let us represent those members by $x$ and $y$; then we have $x + y = 10$; and $x = 10 - y$, where $y$ is so far only determined, that this letter must represent an integer and positive number. We may therefore substitute for it all integer numbers from 1 to infinity: but since $x$ must likewise be a positive number, it follows, that $y$ cannot be taken greater than 10, for otherwise $x$ would become negative; and if we also reject the value of $x = 0$, we cannot make $y$ greater than 9; so that only the following solutions can take place: If $y = 1, 2, 3, 4, 5, 6, 7, 8, 9$, then $x = 9, 8, 7, 6, 5, 4, 3, 2, 1$. But, the last four of these nine solutions being the same as the first four, it is evident, that the question really admits only of five different solutions. If three numbers were required, the sum of which might make 10, we should have only to divide one of the numbers already found into two parts, by which means we should obtain a greater number of solutions. 4. As we have found no difficulty in this question, we will proceed to others, which require different considerations. Question 1. Let it be required to divide 25 into two parts, the one of which may be divisible by 2, and the other by 3. Let one of the parts sought be $2x$, and the other $3y$; we shall then have $2x + 3y = 25$; consequently, $2x = 25 - 3y$; and, dividing by 2, we obtain $x = \frac{25-3y}{2}$; whence we conclude, in the first place, that $3y$ must be less than 25, and, consequently, that $y$ is less than 8. Also, if, from this value of $x$, we take out as many integers as we possibly can, that is to say, if we divide by the denominator 2, we shall have $x = 12 - y + \frac{1-y}{2}$; whence it follows, that $1 - y$, or rather $y - 1$, must be divisible