<s>[INST] Probability that the product of two numbers in $\{0,1,...,10^{n - 1}\}$ contains $2n - k$ digits, $k = 0,1,...$ How to find the probability that the product of two numbers in ${0,1,...,10^n - 1}$ contains $2n-k$ digits, $k = 0,1,...$? As I understand, the sum of all the combinations of two numbers is equal to $10^{2n}$. But I can't understand how to find the sum of all the positive combinations. I understand that the product of these numbers is in $10^{2n-k-1} \le x*y < 10^{2n-k}$. But it doesn't help me. Explain me please, how I can find the sum of all the positive combinations.
 [/INST] This will give an approximate answer, which will be very close.  The region of interest is the square $[0,10^n-1] \times [0,10^n-1]$ in the $xy$ plane.  The numbers with $2n-k$ digits are the numbers from $10^{2n-k-1}$ to $10^{2n-k}-1$.  Instead of counting lattice points, we can view $x$ and $y$ as continuous variables and find the area that corresponds to this range of products.  This is where the approximation comes from.  It is the area between the two hyperbolas $xy=10^{2n-k-1}$ and $10^{2n-k}$.  Do the integral and you are done.
Added:  let us do an example with $n=7, k=4$ so we are looking for factors less than or equal to $10^6$ that multiply to an eight digit number.  Such numbers are in the range $[10^7,10^8)$  Our square is $10^6 \times 10^6$ for a total area of $10^{12}$.  The lower boundary of the area we are looking at is $xy=10^7$ and the upper boundary is $xy=10^8$.  The lower hyperbola hits the edges of the square at $(10,10^6)$ and $(10^6,10)$.  The area below it and inside the square is $$10 \cdot 10^6+\int_{10}^{10^6}\frac {10^7}x\ dx=10^7+10^7 \log(10^5) \approx 1.2513\cdot 10^8$$
where the first term is the rectangle to the left of the intersection with the top of the square.  Similarly the upper hyperbola hits the square at $(100,10^6)$ and $(10^6,100)$ so the area below it is
$$100 \cdot 10^6+\int_{100}^{10^6}\frac {10^8}x\ dx=10^8+10^8 \log(10^4) \approx 10.2103\cdot 10^8$$
The difference of areas is then $8.959 \cdot 10^8$  That is approximately the number of pairs that have a product with eight digits.
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