Number of integers of the form $2^{\alpha}\cdot 5^{\alpha_1} < n$ Let $n$ be a fix integer. 
I would like to find the number of integers of the form : $2^{\alpha} \cdot 5^{\alpha_1} < n , (\alpha, \alpha_1) \in \mathbb{N}^2$.
Here is what I've done so far : 
For the integers of the form : $2^{\alpha}$, it's clear that there are : $[\log_2(n)]$ that are $ <n$.
For the integers of the forme $5^{\alpha_1}$, it's clear that there are : 
$[\log_5(n)]$ that are $< n$.
Yet I don't knwo how to extend this way of thinking for the integers of the form : $2^{\alpha} \cdot 5^{\alpha_1}$...
EDIT : maybe something like : $[\log_{10}([n/5])]$ ? 
 A: Fix $\alpha$ and count the number of $\alpha_1$:
$$\alpha= 0 \Rightarrow \alpha_1 \leq \left[\log_5(n)\right]$$
$$\alpha= 1 \Rightarrow \alpha_1 \leq \left[\log_5\left(\frac{n}{2}\right)\right]$$
$$\cdots$$
$$\alpha = \left[\log_2\left(n\right)\right] \Rightarrow \alpha_1 \leq \left[\log_5\left(\frac{n}{2^{\left[\log_2\left(n\right)\right]}}\right)\right]$$
Hence the number of all pairs would be:
$$\sum_{i=0}^{\left[\log_2\left(n\right)\right]}\left[\log_5\left(\frac{n}{2^i}\right)\right] < \sum_{i=0}^{\left[\log_2\left(n\right)\right]}\log_5\left(\frac{n}{2^i}\right) = \log_5\left(\frac{n^{[\log_2(n)]}}{2^{\frac{[\log_2(n)]\times ([\log_2(n)]+1)}{2}}}\right) \approx \log_5(n^{\frac{[\log_2(n)]}{2}}) = (\frac{[\log_2(n)]}{2})\times\log_5(n) \approx \left[\frac{\left[\log_2(n)\right]\times \log_5(n)}{2}\right]$$
A: It is clearer if you take logarithms.  You are asking that $\alpha \log 2 + \alpha_1 \log 5 \lt \log n$  You can visualize this as a lattice with spacing $\log 2$ in the horizontal direction and $\log 5$ in the vertical direction.  The numbers of interest are the lattice points lying below the line $\alpha \log 2 + \alpha_1 \log 5 = \log n$  The area below the line is $\frac 12(\log n)^2$ and each block of the lattice is area $\log 2 \cdot \log 5$ so the expected number is $\frac {(\log n)^2}{2 \log 2 \cdot \log 5}$  The $+1$s in Raffaele's answer come from the fact that the points of interest are the lower left corner of the blocks.
A: Starting from OmG formula I found this one which is far more precise
$$\left\lfloor \frac{1}{2} \left(\log _2(n)+1\right) \left(\log _5(n)+1\right)\right\rfloor$$
Don't know why but it gives almost exact values
For instance, for $n = 1234567890$ counting the actual numbers I found $218$ while the formula found by OmG gives about $196$ while the "enhanced" formula gives $218$.
Try it for $100!$ it gives the exact result $59\,675$...
