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I just want to add some additional information to the answer of Ilmari. As Ilmari has already described in his answer, when using RSA you work in the ring of integers ${\mathbb Z}/{\mathbb Z}_n$, which is also called a residue class ring. This means that it consists of the set of residue classes $[i]$, where the $i$'th class is defined as the set $\{a ... 6 As mikeazo notes in the comments, RSA operates on the ring$\mathbb Z / n\mathbb Z$of integers modulo$n$, for a given modulus$n = pq$. In this ring, $$E(m) \cdot t^e \equiv m^e \cdot t^e \equiv (mt)^e \equiv E(mt)\ \pmod n.$$ In particular, for$n = 35$,$e = 23$,$$17^{23} \cdot 2^{23} \equiv 33 \cdot 18 \equiv 594 \equiv 34 \equiv 34^{23}\ ... 4 Yes. Assume that the attacker knows the ciphertext$c = c_1 \mathbin\| c_2$, the initialization vector$v$and the plaintext$m = m_1 \mathbin\| m_2$. This tells them that$D_k(c_1) = m_1 \oplus v$and$D_k(c_2) = m_2 \oplus c_1$, where$D_k(\cdot)$denotes block cipher decryption under the (unknown) key$k\$. In particular, this implies that, if the ...