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238 CHAPTER 12. RSA
Neither of these processes is very hard, but each is a bit time-consuming,
especially because there are several things we wrote again and again, like ÷2
and (mod 171). Fortunately, we can combine the two steps and cut out much
of the unnecessary writing. As a first example we redo the previous one.
Example: Compute 23 159 %171.
exponent quotient remainder power of 2 modulo 171
159 ÷2 79.5 1 1 23 ≡ 23
79 ÷2 39.54 1 2 23 2 ≡ 16
39 ÷2 19.5 1 4 16 2 ≡ 85
19 ÷2 9.5 1 8 85 2 ≡ 43
9 ÷2 4.5 1 16 43 2 ≡ 139
4 ÷2 2 0 32 139 2 ≡ 169
2 ÷2 1 0 64 169 2 ≡ 4
1 ÷2 .5 1 128 4 2 ≡ 16
Thus 23 159 ≡ 16 · 43 · 85 · 16 · 23 ≡ 163 (mod 171).
⋄
From now on we will abbreviate by not writing the “÷2 =” or “power”
columns. 5
Example: Compute 45 61 %79.
First, 45 < 79 and 61 < 70 so the base and the power are already reduced.
Since 45 61 is far too big we must build we will call the binary chart.
exponent quotient remainder mod 79
61 30.5 1 45 ≡ 45
30 15 0 45 2 ≡ 50
15 7.5 1 50 2 ≡ 51
7 3.5 1 51 2 ≡ 73
3 1.5 1 73 3 ≡ 36
1 .5 1 36 2 ≡ 32
Thus 45 61 ≡ 45 · 51 · 73 · 36 · 32 ≡ 2 (mod 79).
⋄
12.4 Complication III: a mini one
The computation we just did, 45 · 51 · 73 · 36 · 32 ≡ 2 (mod 79), almost is too
much for many calculators. It certainly can happen that the final computation’s
product is too big. Then what If there are too many factors, or if they are too
5 In fact, one of the interesting results of this method is that we don’t need to explicitly
find the proper powers of 2.