Then 3 * x+4 * y+1/7 * z =100.

x+y+z= 100

The three unknowns of the two equations have numerous solutions, but x, y and z must be integers with their own ranges.

For the first equation, z must still be a multiple of 7, so z starts from 7 and the value is 7, 14, 2 1, ... 9 1.

It is concluded that there are only unique solutions x=3, y=20 and z=77.