Number of ordered pairs $(a,b)$ such that $ab \le 3600$ Find number of ordered pairs $(a,b)$ such that $ab\le 3600$ and $a,b \in N$
My attempt : Well, all $a,b \leqslant 60$ are solutions. These 3600 solutions. After that I have no idea how to count the number of remaining solutions... 
I have a feeling that my approach is $very$ bad. But I had to give as it would be marked as off-topic.
 A: Hint:  First, you can count only those pairs where $a \lt b$, double them, and add one for $(60,60)$  We know $a \lt 60$.  How many are there for $a=1$?  How many are there for $a=2$?  Can you write an expression for general $a$?  Then add them up for $a$ from $1$ to $59$.
A: You ask for the count of ordered pairs of integers (a,b) such that a*b <= n.
Set up a table in which the first row fixes a to 1: this allows b to run from 1 to n.
In the k'th row, we fix a to value k, so b runs from 1 to floor(n/k).
The table is obviously symmetrical across the diagonal (1,1), .. (q,q) with q=floor(sqrt(n))). This allows us to simplify the counting to a running from 1 to q, take twice that, and subtract the double counting of the square q^2.
Using Mathematica:
Table[q = Floor[Sqrt[n]]; 2*Sum[Floor[n/k], {k, q}] - q^2, {n, 16}]
or {1,3,5,8,10,14,16,20,23,27,29,35,37,41,45,50}
But, looking it up produces http://oeis.org/A006218 :
Table[ Sum[DivisorSigma[0,k],{k,n} ],{n,16}]  normally written as $\sum _k^n \sigma _0(k)$ alias the sum of (the count of divisors of k) for k running from 1 to n.
The challenge is to prove both counts to be equal.
Looking down the columns of the above table, we see that the entry in row k is 1 larger than that in row k-1 if floor(n/k) > floor(n/(k-1)), and that again happens only if k is a divisor of n. So, within row k, the entries in column j are 1 larger than in row k-1 exactly if j divides n. Not very challenging after all.
