GK 2015 Q24

Suppose we obtain the remainder \(r\) when we divide 2015 by the positive integer \(d\). Then \(r \leqslant d - 1\). Also, with \(d \leqslant 1000\) the quotient must be at least 2. This suggests that to get the largest possible remainder we should aim to write 2015 in the form \(2d + (d - 1)\). The equation \(2015 = 2d + (d - 1)\) has the solution \(d = 672\). So we obtain the remainder 671 when we divide 2015 by 672. If we divide 2015 by an integer \(d_1 < 672\), the remainder will be at most \(d_1 - 1\) and so will be less than 671. If we divide 2015 by an integer \(d_2\), where \(672 < d_2 \leqslant 1000\) and obtain remainder \(r_2\), we would have \(r_2 = 2015 - 2d_2 < 2015 - 2 \times 672 = 671\). Hence 671 is the largest remainder we can obtain.