Let x, y, and z be the three quotients of the divisions; the number sought will then be expressed by 28 x + 10, by 19 y + 2, or by 15 z + 4. Hence the two equations
28 x + 10 = 19 y + 2 = 15 z + 4.
| To solve the equations 28 x + 10 = 19 y + 2, or y = x + | 9 x + 8 19 | , let m = | 9 x + 8 19 | , we have then x = 2 m + | m - 8 9 | . |
| Let | m - 8 9 | = m′; then m = 9 m′ + 8; hence |
x = 18 m′ + 16 + m′ = 19 m′ + 16 . . . (1).
Again, since 28 x + 10 = 15 z + 4, we have
| 15 z = 28 x + 6, or z = 2 x - | 2 x - 6 15 | . |
| Let | 2 x - 6 15 | = n; then 2 x = 15 n + 6, and x = 7 n + 3 + | n 2 | . |
| Let | n 2 | = n′; then n = 2 n′; consequently |
x = 14 n′ + 3 + n′ = 15 n′ + 3 . . . (2).