maze's question is classic... the answer is

Spoiler

N-floor (sqrt(N))

the nice animation makes it kinda obvious... the reason is only perfect squares have an odd number of distinct divisors.

problem 2 above

Spoiler

it suffices to find # of ways to arrange coins such that each row has one coin. This is easily given as [tex]n^n[/tex], so answer is [tex]1-n^n/ \binom{n^2}{n}[/tex]... if you want to find probability such that at least one row or one column has no silver coin, then it would be [tex]1-n!/ \binom{n^2}{n}[/tex]

problem 3 above

Spoiler

LHS is less than or equal to [tex]a(a+b+c)+b(a+b+c)+c(a+b+c)[/tex] by weighted AM-GM inequality, which is clearly less than or equal to 1

Problem 1 seems too long and I am feeling kinda lazy...

Now for my problem... hmmmm... after all these years.... let's see.....I'm not sure if this is appropriate:

prove that all one dimensional compact connected Lie group (a manifold that is also a group with smooth multiplication) is diffeo-isomorphic to S1 (the circle).

In case we wanna make things "elementary", evaluate:

[tex]\int_{0}^\infty \frac{dx}{1+x^n}[/tex]

In case you think a simple contour will bring this problem down... try to evaluate it for n>1, not just integers.

edit: sorry, latex just doesn't work well with spoiler alerts...