Download Fast modular exponentiation, or, how to compute residues of

Fast modular exponentiation, or, how to compute residues of
numbers bigger than the number of atoms in the universe
Suppose you’re given a large power of some number, like 5321 , and some
other number, like 123. Is it possible to quickly find the remainder r of 5321
when it is divided by 123? Note that if you ask Wolfram Alpha, it says the
number of atoms in the universe is thought to be about 1080 . So how can we
possibly expect to compute 5321 and divide it by 123?
The trick is that we don’t have to compute the full decimal expansion of 5321
in order to find r. Instead, we can exploit the arithmetic of exponents, and the
fact that every remainder of any number divided by 123 is between 0 and 123.
Doing so will keep us from ever having to compute a number bigger than 1232 .
In detail, here is how this works. First, we write 321 as a sum of powers of
2. For convenience, I will start by writing down all powers of 2 less than 321:
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
28 = 256
29 = 512
We can write
321 = 256 + 64 + 1.
(Basically, we are finding the binary expansion of 321. Indeed, if we rewrite the
above equation as
321 = 1 · 256 + 0 · 128 + 1 · 64 + 0 · 32 + 0 · 16 + 0 · 8 + 0 · 4 + 0 · 2 + 1 · 1
= 1 · 28 + 0 · 27 + 1 · 26 + 0 · 25 + 0 · 24 + 0 · 23 + 0 · 22 + 0 · 21 + 1 · 20
then we see that we can write 321 as 101000001 in binary.) Now we observe
that
5321 ≡ 5256 · 564 · 51 mod 123.
Thus, if we can compute 5256 mod 123, 564 mod 123, and 51 mod 123, we can
find 5321 mod 123. Now obviously
0
52 ≡ 51 ≡ 5
1
mod 123,
and
1
52 ≡ 52 ≡ 25
mod 123.
But instead of writing
2
52 ≡ 54 ≡ 625
mod 123,
and leaving it at that, we can use long division to reduce, and write
2
52 ≡ 10
mod 123.
3
This pays off when we move to computing the remainder of 52 when we divide
it by 123, since we can write
3
2
52 = 52
+22
2
2
= 52 52 ,
and, hence,
3
2
2
52 ≡ 52 52 ≡ 10 · 10 ≡ 100
mod 123.
Hopefully you’ll agree with me that finding 10 · 10 mod 123 is a lot easier to do
3
than finding 52 ≡ 58 ≡ 390625 mod 123.
8
Now we do this trick repeatedly until we have found 52 mod 123. For
example, we compute
4
3
3
52 ≡ 52 52 ≡ 1002 ≡ 10000
mod 123.
Now, using long division, we see that
10000 = 81 · 123 + 37,
so
4
52 ≡ 37
mod 123.
2
Continuing in this way, 37 = 1369, and long division shows
1369 = 11 · 123 + 16
so
5
52 ≡ 1369 ≡ 16
mod 123.
And so on (I am just going to write the answers, but let me reiterate that
all of this can be done easily by hand using long division):
6
52 ≡ 10
mod 123,
7
52 ≡ 100
mod 123,
8
52 ≡ 37
mod 123.
Now we put together what we need to, and use long division one more time, to
conclude
8
6
0
5321 ≡ 5256 · 564 · 51 ≡ 52 · 52 · 52 ≡ 37 · 10 · 5 ≡ 1850 ≡ 5
2
mod 123.