Chapter 2 - Exercises
1. Among the shifts of EVIRE, there are two words: arena and river. Therefore,
Anthony cannot determine where to meet Caesar.
2. The inverse of 9 mod 26 is 3. Therefore, the decryption function is
x = 3(y −2) = 3y −2 (mod 26). Now simply decrypt letter by letter as follows.
U = 20 so decrypt U by calculating 3 ∗ 20 − 6 (mod 26) = 2, and so on. The
decrypted message is ’cat’.
3. Changing the plaintext to numbers yields 7, 14, 22, 0, 17, 4, 24, 14, 20.
Applying 5x+7 to each yields 5 ·7 + 7 = 42 ≡ 16 (mod 26), 5 ·14 + 7 = 77 ≡ 25,
etc. Changing back to letters yields QZNHOBXZD as the ciphertext.
4. Let mx + n be the encryption function. Since h = 7 and N = 13, we
have m · 7 + n ≡ 13 (mod 26). Using the second letters yields m · 0 + n ≡ 14.
Therefore n = 14. The first congruence now yields 7m ≡ −1 (mod 26). This
yields m = 11. The encryption function is therefore 11x + 14.
5. Let the decryption function be x = ay + b. The firs t letters tell us that
7 ≡ a ·2 + b (mod 26). The second letters tell us that 0 ≡ a ·17 + b.Subtracting
yields 7 ≡ a · (−15) ≡ 11a. Since 11
−1
≡ 19 (mod 26), we have a ≡ 19 · 7 ≡ 3
(mod 26). The first congruence now tells us that 7 ≡ 3 · 2 + b, so b = 1. The
decryption function is therefore x ≡ 3y + 1. Applying this to CRWWZ yields
happy for the plaintext.
6. Let mx+n be one affine function and ax+b be another. Applying the first
then the second yields the function a(mx + n) + b = (am)x + (an + b), which is
an affine function. Therefore, successively encrypting with two affine functions
is the same as encr ypting with a single affine function. There is therefore no
advantage of doing double encryption in this case. (Technical point: Since
gcd(a, 26) = 1 and gcd(m, 26) = 1, it follows that gcd(am, 26) = 1, so the affine
function we obtained is still of the required form.)
7. For an affine cipher mx + n (mod 27), we must have gcd(27, m) = 1,
and we can always take 1 ≤ m ≤ 27. So we must exclude all multiples of 3,
which leaves 18 possibilities for m. All 27 values of n are possible, so we have
18 · 27 = 486 keys. When we work mod 29, all values 1 ≤ m ≤ 28 are allowed,
so we have 28 · 29 = 812 keys.
8. (a) In order for α to be valid and lead to a decryption algorithm, we need
gcd(α, 30) = 1. The possible values for α are 1, 7, 11, 13, 17, 19, 23, 29.
(b) We need to find two x such that 10x (mo d 30) gives the same value.
There are many such possible answers, for example x = 1 and x = 4 will work.
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