A hundred loaves for five men

Question

Divide 100 loaves among five men so that the five shares form an arithmetic progression (each share exceeds the previous by the same amount) and the sum of the three largest shares is seven times the sum of the two smallest. Find the five shares. This is problem 40 of the Rhind Mathematical Papyrus, written around 1550 BC.

Accepted Answer

The shares are 1 2/3, 10 5/6, 20, 29 1/6 and 38 1/3 loaves. The middle share. With shares a minus 2d, a minus d, a, a + d, a + 2d, the total is 5a = 100, so a = 20. The ratio condition. The three largest shares total a + (a + d) + (a + 2d) = 3a + 3d = 60 + 3d. The two smallest total (a minus 2d) + (a minus d) = 2a minus 3d = 40 minus 3d. The condition says 60 + 3d = 7(40 minus 3d) = 280 minus 21d, so 24d = 220, giving d = 55/6 = 9 1/6. The shares. 20 minus 2d = 20 minus 18 1/3 = 1 2/3; 20 minus d = 10 5/6; 20; 20 + d = 29 1/6; 20 + 2d = 38 1/3. Check. The total is 100. The three largest: 20 +…

A hundred loaves for five men

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