TITLE: How many order triple (a,b,c) are there such that $a.b.c \le 1000$
QUESTION [5 upvotes]: How many order triple of positive integers (a,b,c) are there such that $a\leq b\leq c$ and $a\cdot b \cdot c \leq 1000$?
I have no idea how to attack this type of problem. Can anyone help. Thanks in advance.

REPLY [4 votes]: In general, the number of positive integer solutions to $abc\le N$ with $a\le b\le c$ can be computed as
$$C(N)=\sum_{a=1}^{\lfloor\sqrt[3]{N}\rfloor}\sum_{b=a}^{\lfloor\sqrt{N/a}\rfloor}(\lfloor N/ab\rfloor-b+1)$$
Alternatively, let
$$T=\lfloor\sqrt[3]N\rfloor,\quad 
D=\sum_{n=1}^\sqrt{N}\left\lfloor{N\over n^2}\right\rfloor,
\quad\text{and}\quad
S=\sum_{n=1}^N\sigma(n)\left\lfloor{N\over n}\right\rfloor$$
where $\sigma(n)$ is the number of divisors of $n$. Then
$$C(N)={S+3D+2T\over6}$$
The latter formula is obtained by counting unordered triples.  Basically, $T$ counts the number of triples $(a,a,a)$ with $a^3\le N$, $D$ counts the number of triples $(a,a,c)$ with $a^2c\le N$, and $S$ counts the number of triples $(a,b,c)$ with $abc\le N$: that is, there are $\sigma(n)$ pairs $(a,b)$ such that $ab=n$, and for each such pair there are $\lfloor N/n\rfloor$ possible values of $c$.  
To give a very simple example, if $N=4$, then $T=1$, $D=4+1=5$, and 
$$S=1\cdot4+2\cdot2+2\cdot1+3\cdot1=4+4+2+3=13,$$ 
so 
$$C(4)={13+3\cdot5+2\cdot1\over6}={13+15+2\over6}=5,$$ which agrees with the explicit list of ordered triples $(1,1,1),(1,1,2),(1,1,3),(1,1,4)$, and $(1,2,2)$.  For a number like $N=1000$, however, the first, nested-sum formula is probably easier to compute, since there are only a few hundred terms altogether and you don't have to compute $\sigma(n)$ for a thousand values of $n$.