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236 CHAPTER 12. RSA
(2) Compute 22 57 %61 (with fewer explanations).
We have 57 = 32 + 16 + 8 + 1. Computing the powers,
22 2 ≡ 15 (mod 61),
22 4 = (22 2 ) 2 ≡ 15 2 ≡ 42 (mod 61),
22 8 = (22 4 ) 2 ≡ (42) 2 ≡ 56 (mod 61),
22 16 = (22 8 ) 2 ≡ (56) 2 ≡ 25 (mod 61), and
22 32 = (22 16 ) 2 ≡ (25) 2 ≡ 15 (mod 61).
Using the powers that we need gives
22 57 = 22 32 · 22 16 · 22 8 · 22 ≡ 15 · 25 · 56 · 22 ≡ 47 (mod 61).
Therefore 22 57 %61 = 47.
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄
In each of these examples we broke down the exponent into a sum of power
of 2. 4 To make this process easier we need to find a way of determining which
powers of 2 are needed. A simple on is as follows. At each step divide by 2. If
this division has a remainder (the quotient has a .5 as a decimal) then indicate
this with a 1, otherwise indicate this with a 0. Then repeat starting with the
integer part of the quotient. End when the final quotient is .5, as it has no
integer part to continue with.
Examples: Find the needed powers for an exponent of 27 and 57.
(1) For 27:
exponent quotient remainder power of 2
27 ÷2 = 13.5 1 1
13 ÷2 = 6.5 1 2
6 ÷2 = 3 0 4
3 ÷2 = 1.5 1 8
1 ÷2 = .5 1 16
So 27 as a 16, 8, 2 and 1 in it (that is 27 = 16 + 8 + 2 + 1).
4 Why 2 and not, say, 3 Because every number can be expressed as a sum of powers of 2,
allowing us to compute 22 57 %61 by performing simply a number of squarings. If we had used
3 instead we would have needed to write 57 = 2 · 27 + 3, a sum of multiples of powers of 3.
Using 2 means we don’t need multiples, which makes the calculations a bit easier.