Very hard one, dont even try it :D. The real question is at the end if this post.

'Euler and Gauss meet in hell. Lucifer promises to let them go if they guess the two numbers between 1-100 he was thinking about. Gauss get the product and Euler the sum of the two numbers. Heres the dialog:
G: I dont know the two numbers.
E: I knew that!
G: I know them now.
E: Me too.'

Which two numbers did they get?

The mother is 27.75 years old and the child is 6.75 years old . . . I don't get it.

Let 'c' = child, 'm' = mother.
The two equations we have are:
c + 21 = m
5*(c + 6) = m + 6

substituting (1) into (2), we get

5(c+6) = c + 27

which gives us
4c = -3
or
c = -3/4

In other words, the child is -3/4 years old, or 9 months. And what happens 9 months before a child is born?

Here's a more elementary proof. It uses only the pigeonhole principle and the unique factorization theorem.

Assume that none of the integers in the set Sn = {10^1-1, ..., 10^n-1} is divisible by n, where n is not divisible by 2 or 5. By the pigeonhole principle, at least two of these integers must have the same remainder when divided by n -- there are n integers in the set, and only n-1 distinct remainders available since 0 has been ruled out. Let 10^j-1, 10^k-1 be one such pair, with n >= j > k > 0. Then n divides their difference, (10^j-1) - (10^k-1) = 10^j - 10^k = (10^(j-k) - 1)(10^k). That is, there exists an integer m such that mn = (10^(j-k) - 1)(10^k).

Consider the prime factorization of each side of this equation. By the unique factorization theorem, each of the 2's and 5's that make up the factorization of the 10k on the right must appear in the factorization of either n or m on the left. But we are given that n is not divisible by 2 or 5, so all the factors must come from m. Thus we know that m' = m/10k is an integer, and m'n = 10^(j-k) - 1. But j-k < n, so 10^(j-k) - 1 is an element of Sn. This contradicts our assumption that no such numbers are divisible by n.

Therefore there exists an element of Sn that is divisible by n.

Not sure I understand this one either. If the train going east arrives, say, 1 minute before the trains going west, then Bill will, on average, end up with Monica 9 times more often than with Hillary. But this solution depended on the arrival times of the trains relative to each other , of which you gave no information.

Indeed! I had 5c = m + 6 !! Stupid mistake, my bad. Nice riddle

Sum of the digits