Maximum length in bits of the product n=pq If p and q are two 1024-bit prime numbers, would the maximum length in bits of product $n=pq$ be ${1024}^2$ or $2^{1024}$?

I think it is ${1048576}={1024}^2$?

Also, would the least significant bit of $n$ be a $0$ or a $1$? I am assuming it would be $0$? Any explanation of what determines significance would be great
If $p$ and $q$ are 1024-bit primes, then by definition of the bit size of an integer (at least, a prime in a cryptographic context with glimpses of RSA), $2^{1023}\le p&lt;2^{1024}$ and $2^{1023}\le q&lt;2^{1024}$. Thus their product $n=pq$ verifies $2^{2046}\le n&lt;2^{2048}$, and $n$ is a 2047-bit or 2048-bit integer. We show by exhibition that both cases are possible, with $p=q=2^{1023}+1155$ for the first case, and $p=q=2^{1024}-105$ for the second.

Hence the maximum length in bits of the product $n=pq$ is $2048=2\cdot1024$.

This generalizes to: for odd primes $p$ and $q$, the maximum length in bits of the product $n=pq$ is the sum of the bit length of primes $p$ and $q$; proof that this bound is tight can be obtained from the observation that $\forall k&gt;1,\exists p\text{ prime with }2^{k-1/2}&lt;p&lt;2^k$, which itself follows from stronger versions of Bertrand's postulate.

Since $p$ and $q$ are large primes, they are odd, hence $pq$ is odd, hence the least significant bit of $n=pq$ is 1.

Addition: in common RSA key generation standards FIPS&nbsp;186-4 appendix B.3 and ancestor ANSI&nbsp;X9.31-1998, it is chosen primes $p$ and $q$ with $2^{k-1/2}&lt;p&lt;2^k$ and $2^{k-1/2}&lt;q&lt;2^k$, which insures that $pq$ is exactly $2k$ bits.