How to learn the first bit of m in padded RSA Consider the following version of padded RSA encryption, where encryption of $m$ is done by setting $m′ = (0^k~||~r~||~00000000~||~m)$ for a random $r$ (of length 8 btyes = 64 bits) and then computing the ciphertext $c = (m′)^e$ (mod $N$). For concreteness, further assume that $N$ is a 1024-bit RSA modulus, and that the scheme only handles 512-bit messages.
For the rest of the problem, assume you are given some ciphertext $c$ that is the encryption of some 512-bit message.
Suppose you learn that $(2^e)c$ (mod $N$) is a valid ciphertext. What does this tell you about the first bit of $m$?
I assume that this is a homework problem, and so I won't give you the answer.  However, here are a few hints:


What does 'it is a valid ciphertext' mean?
If $c$ decrypts to the value $m'$ (not counting the depad operation; that is, if $m' = c^{1/e} \bmod N\ $), what does $2^ec \bmod N\ $ decrypt to?  What is the relationship to those two values?