Elements of Algebra

306 ELEMENTS PART II. sought $N = 78z - 10$; therefore, the question admits of the following values of N; viz. $N = 68, 146, 224, 302, 380, \ldots$ which numbers form an arithmetical progression, whose difference is $78 = 6 \times 13$. So that if we know one of the values, we may easily find all the rest; for we have only to add 78 continually, or to subtract that number, as long as it is possible, when we seek for smaller numbers. 14. The following question furnishes an example of a longer and more tedious solution. Question 8. To find a number N, which, when divided by 39, leaves the remainder 16; and such also, that if it be divided by 56, the remainder may be 27. In the first place, we have $N = 39p + 16$; and in the second, $N = 56q + 27$; so that $39p + 16 = 56q + 27$, or $39p = 56q + 11$, and $p = \frac{56q + 11}{39} = q + \frac{17q + 11}{39} = q + r$, by making $r = \frac{17q + 11}{39}$. So that $39r = 17q + 11$, and $q = \frac{39r - 11}{17} = 2r + \frac{5r - 11}{17} = 2r + s$, by making $s = \frac{5r - 11}{17}$, or $17s = 5r - 11$; whence we get $r = \frac{17s + 11}{5} = 3s + \frac{2s + 11}{5} = 3s + t$, by making $t = \frac{2s + 11}{5}$, or $5t = 2s + 11$; whence we find $s = \frac{5t - 11}{2} = 2t + \frac{t - 11}{2} = 2t + u$, by making $u = \frac{t - 11}{2}$; whence $t = 2u + 11$. Having now no longer any fractions, we may take u at pleasure, and then we have only to trace back the following values: $t = 2u + 11$, $s = 2t + u = 5u + 22$, $r = 3s + t = 17u + 77$, $q = 2r + s = 39u + 176$, $p = q + r = 56u + 253$,