RSA encryption using multiplication Generally in RSA we encrypt as $m^e \pmod n$. Will RSA work if we replace the power by normal multiplication? $E = (m \times e) \mod n$  and decryption as $c \times d \mod n$. What will be $d$ disadvantage if it works ?
Suppose an RSA public key is still $(e, N)$ and the private key is still $d$. We have $Enc = (m*e) \text{ mod } N$ and $Dec = (c*d) \text{ mod } N$. The correctness of the scheme depends on the fact that $Dec(Enc(m)) = (m*e*d) \text{ mod } N = m \text{ mod } N$. This implies $e*d = 1 \text{ mod } N$. It is thus trivial to compute the private key given only the public key (i.e. compute $d = e^{-1} \text{ mod } N$). Hence this scheme is completely broken.

Note that this scheme only works if $e$ and $N$ are coprime otherwise no $d$ such that $e*d = 1 \text{ mod } N$ exists.