Code and ciphers: Julius Caesar, the Enigma and the internet

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appendix
so
but
so
Hence
1�6�1�(17�2�6)�3�6�17
6�91�5�17
1�3�(91�5�17)�17�3�91�16�17.
m�3 and n�16.
(Check: 3�91�273 and 16�17�272.)
An alternative method
The values of m and n can also be found by using continued fractions [6.7].
Although it looks different the method is essentially the same as that of
the Euclidean Algorithm.
To illustrate the method we again find the values of m and n such that
91m�17n�1.
91 6
�5� ,
17 17
17 5
�2� ,
6 6
6 1
�1� .
5 5
The partial quotients of the continued fraction are therefore (5, 2, 1, 5) and
its convergents are
5 11 16 91
, , , and .
1 2 3 17
The numbers m and n are the numerator and denominator of the penultimate
convergent: 16 and 3, as we found before.
M26 Efficiency of finding powers by repeated squaring
Given a number, X, that we wish to raise to the power n we could compute
X n by multiplying X by itself (n�1) times. If n is small this is reasonable
but if n is large it is very inefficient. Let k be such that
2 k �n�2 (k�1) ;