Greta sent us her work on this problem:

Since he has friends older than he is, I think it's reasonable to assume that his 100th birthday wasn't before 1992 (as otherwise his friends would be older than 110).

Now D = 2pqr, where p, q, r are distinct primes (and not 2).
Also, D must be one of 1992, 1994, 1996, ..., 2002, so pqr is one of 996, 997, 998, ..., 1001.
But pqr is odd, so pqr is one of 997, 999, 1001.
Now 997 is prime and 999 is divisible by 9, so it can't be either of those.
But 1001 = 7x11x13, so this works. So D = 2002, so A = 1902.
We can check that this satisfies the other requirements of the question:

A = 1902 = 2x3x317
B = 1922 = 2x31x31
C = 1982 = 2x991