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:
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.