(b) What is the maximum number of ASCII letters that can be encrypted in one go and sent to Alice (assuming the message has no padding)? Explain your answer. (c) Alice decides to change the e value in her public key. Which of the following choices are not allowed: 3,5,7,13,17,19,35,45? Explain your answer.
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b) The answer to this part may depend on what ASCII characters you mean. The upper limit is 7 bits per character and includes numerals punctuation and other signs. 7 bits have a range of 128 values. If n is the number of characters, then 128ⁿ must be less than the modulo, which is 62663. 128²=16384. But 128³=2097152, which is bigger than 62663. So the maximum on this basis is 2. This is my best guess.

If the ASCII codes can only be upper- and lowercase letters then, since there are 52 of these, 52²=2704, but 52³=140608, which is bigger than 62663. So, even with this reduction, only 2 letters can be sent in one go. So the answer is 2, however the characters are encrypted.

Another way to answer this is to count bits. ASCII codes are 7 bits long and N is 16 bits long, so only two will fit. 52 is 66₈ which is 6 bits long, so only two will fit.

c) e cannot be a factor of 222 or 280.

222=2×3×37, so 3 is disallowed.

280=8×5×7, so 5 and 7 are disallowed, and so is 35.

45 is disallowed because it has a factor 5 in common with 280.

I always intend to answer all parts of your questions, if I can, but to do so I need to research the subject more, since cryptography is not a subject I have studied. Also, I have great difficulty submitting answers and editing, because the system and/or the network is not working properly on this website. It has been like this for many months. In particular, editing becomes nearly impossible for lengthy solutions and I experience frequent system crashes or loss of response, extreme sluggishness, reloads, etc., so it can take a long time to produce and submit a complete answer.

In the meantime, other users have the opportunity to use their skills and expertise to answer your questions. Maybe they have a better system response than I have.

I am an ordinary user and, apart from some admin functions granted to me, I have no connection with the website and I’m certainly not a mathematics expert. (I just like trying to solve problems and to help people.) So I research and check any solutions for accuracy,  conciseness and presentation to try to help those who submit questions.

I hope you will be patient as I attempt to answer your questions.

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