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Given 2 binary numbers of n digits. (n can be from 1 to something large) What is the maximum number of digits of their product?

I think it's 1 in 1-digit*1-digit, and 2n otherwise. But I want some more proof.

For example, n=3. So I want to know the maximum number of digits in their product. If I used the largest number in 3 digits, 111*111=110001, which is 6 digits. But I want some sort of function that gets the number of digits without actually calculating this.

Kung Pao Chicken
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2 Answers2

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The largest number with $n$ binary digits is $2^n-1$. So the largest possible product is $(2^n-1)^2 = 2^{2n}-2^{n+1}+1$.

If $n=1$, this number is equal to $1$, as you say. Otherwise, it lies strictly between $2^{2n-1}$ and $2^{2n}$, so it has $2n$ digits. (It is greater than $2^{2n-1}$ because $2^{n+1} \le \frac12\cdot 2^{2n}$ if $n \ge 2$.)

TonyK
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  • May I ask what's the relation between $2^{n+1}\leq \frac{1}{2}\cdot 2^{2n}\forall n \geq 2\rightarrow 2^{2n-1}\leq 2^{2n}-2^{n+1}+1$? – Andes Lam Mar 02 '20 at 03:15
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When you square an $n$-digit number you get either $2n-1$ or $2n$ digits in the product: this is true for binary, decimal or any other base. Which one you get depends on whether or not there is a "carry" into the $2n$ column when you make the calculation.

Let your number, $k$, have $n$ binary digits. We want to find how large $k$ must be so that $k^2$ has $2n$ rather than $2n-1$ digits, that is we want to solve

$$2n = \lfloor 2\log_2k+1 \rfloor.$$

$\displaystyle k = \big \lfloor \frac{2^n}{\sqrt 2}\big\rfloor + 1$ is the smallest value that does the job.

For your example, $n=3$ gives $k=6$ and we can check that $6^2 = 36 = 100100_2$ has six binary digits whereas $5^2=25=11001_2$ has only five.

For a larger example, $n=10$ gives $k=725$ and we find that $725^2 = 525625 = 10000000010100111001_2$ has twenty binary digits, whereas $724^2 = 524176 = 1111111111110010000_2$ has a mere nineteen.

Peter Phipps
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  • Thanks for the effort, but maybe you have mistaken or I misled you. :( The number itself does not matter, I just need to know the maximum number of digits from 2 numbers with that digit. A example use would be to write some program which needs to reserve memory for that. I somehow think it's 2n or 1 when the number of digit is 1, but I cannot find the proof, which is what I wanted to know here. I hope this makes it more clear. :) – Kung Pao Chicken Feb 21 '14 at 22:23
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    I don't think that you can avoid using logarithms. Say your $n$-digit numbers have base-$2$ logarithms $\alpha$ and $\beta$ where $n-1 \le\alpha, \beta < n$. Their product will have $2n$ digits if (and only if) $\alpha+\beta > 2n-1$. If you don't actually want to calculate the product I don't suppose you want to calculate the logarithms either. – Peter Phipps Feb 23 '14 at 20:50