Exercise 1: Full RSA Key Generation and Round Trip for p=5, q=11, e=3 — Possible Solution
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STEP 1-2: n
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p = 5, q = 11. n = p*q = 55.

STEP 3: phi(n)
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phi(55) = (5-1)(11-1) = 4*10 = 40

STEP 4: CHECK e=3 IS VALID
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gcd(3, 40): 3 and 40 share no common factors (3 is prime, and 40
is not a multiple of 3) - gcd(3,40) = 1. Valid choice.

STEP 5: COMPUTE d VIA THE EXTENDED EUCLIDEAN ALGORITHM
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Running extended_gcd(3, 40):
40 = 3*13 + 1  -> q=13
3  = 1*3 + 0   -> stop, gcd=1

Back-substituting (or tracking coefficients directly): the
extended Euclidean algorithm gives d = 27.

Check: 3*27 = 81. 81 mod 40 = 1 (since 81 = 40*2 + 1). Confirmed:
e*d ≡ 1 (mod 40).

STEP 6: KEYS
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Public key: (n=55, e=3)
Private key: (n=55, d=27)

ENCRYPT AND DECRYPT m=4
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Encrypt: c = 4^3 mod 55 = 64 mod 55 = 9
Decrypt: m = 9^27 mod 55 = 4

RESULT
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The round trip is confirmed: encrypting 4 gives 9, and decrypting
9 with the private key correctly recovers 4.

WHY THIS WORKS AS AN ANSWER
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Every one of this chapter's own six key-generation steps is carried
out explicitly and in order (including the gcd validity check on e,
not skipped), and the encrypt/decrypt round trip is verified with
real numbers rather than assumed to work because the earlier steps
were correct.