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244 CHAPTER 12. RSA
c = 3 → 3 23 %247 = 243.
o = 15 → 15 23 %247 = 59.
d = 4 → 4 23 %247 = 36.
e = 5 → 5 23 %247 = 47.
So code is send as 243, 59, 36, 47.
2. Decipher 47, 123, 61, 59.
To decipher we must find d. First, (P −1)(Q−1) = 18·12 = 216, and
then from the Euclidean algorithm d = 47 is the solution 23x ≡ 1
(mod 216). Then
47 → 47 47 %247 = e → e.
123 → 123 47 %247 = 24 → x.
61 → 61 47 %247 = 16 → p.
59 → 59 47 %247 = 15 → o.
So the deciphered message is expo.
(2) Use N = 2747 and e = 19 to encipher exponentiation in two-letter pairs.
ex = 0524 → 0514 19 %2747 = 1567.
po = 1615 → 1615 19 %2747 = 1084.
ne = 1405 → 1405 19 %2747 = 1461.
nt = 1420 → 1420 19 %2747 = 1323.
ia = 0901 → 901 19 %2747 = 901.
ti = 2009 → 2009 19 %2747 = 2009.
on = 1514 → 2009 19 %2747 = 1818.
So 1567, 1084, 1461, 1323, 901, 2009, 1818 is the ciphertext.
(3) Use that 2747 = 67 · 41 to to decipher 1032, 1469, 1821, 1551, 2020
into two-letter pairs.
After finding (P − 1)(Q − 1) = 2640 and d = 139, we decipher:
The answer is superpower.
1032 → 1032 139 %2747 = 1921 → su.
1469 → 1469 139 %2747 = 1605 → pe.
1821 → 1821 139 %2747 = 1816 → rp.
1551 → 1551 139 %2747 = 1523 → ow.
2020 → 2020 139 %2747 = 518 → er.
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