The following problem was found in Benjamin Banneker’s only remaining journal:
Divide 60 into four such parts that the first being increased by 4, the second decreased by 4, the third multiplied by 4, the fourth part divided by 4, that the sum, the difference, the product and the quotient shall be one and the same number.
Can you find the four numbers that satisfy these conditions?
Solution:
Let the four numbers be represented by w, x, y, and z. The conditions expressed in the problem are captured in the equations: w+ 4 = x – 4 = 4y = z/4, and w + x + y + z = 60.
w + 4 = x – 4 yields x = W + 8; x – 4 = 4y yields y = (x-4)/4, so y = (w + 4)/4
4y = z/4 yields z = 16 y, so z = 16(w+4)/4 or 4(w +4)
Substituting for w, x, y and z from the equations above, into the equation w + x + y + z = 60. yields:
w + (w +8) + (w + 4)/4+ 4(w+4) = 60 which on simplification becomes 25w = 140. So w = 5 3/5.
Substituting w = 5 3/5 into the equations for w, x, y and z yields: x = 13 3/5, y = 2 2/5, and z = 382/5.
