A, B, C, and D are all different positive integers.
A < B < C < D
Find the largest D so that     1/A + 1/B + 1/C + 1/D = 1


Solution to the Problem:

D = 42

So 1/2 + 1/3 + 1/7 + 1/42 = 1
If D is as large as possible, then 1/D is as small as possible,
and so 1/A, 1/B, and 1/C must be as large as possible.
If A = 2, B = 3, and C = 4, we find that 1/A + 1/B + 1/C > 1.
To find successively smaller values of this sum, we increase the denominator of the
smallest fraction in increments until we have a satisfactory sum.
After seeing that 1/2 + 1/3 + 1/6 = 1,
we try 1/2 + 1/3 + 1/7 = 41/42, which works, because now 1/D = 1/42.


Correctly solved by:

1. James Alarie Flint, Michigan

Partial Credit:
Dawn Chapman   Columbus Technical College, Columbus, Georgia sent in A = 2, B = 4, C = 5, and D = 20.
I believe this is the second largest value of D that works in the equation.