modulo of large numbers (e.g. 13^183 mod 11)
Author
This isn't a homwork question btw, just going through an old maths course to catch up a bit.
Anyway, I know the anser is 8 and done it using simple mod arithmetic.
Here's my working:
13^183 mod 11 =
13^(2^7) mod 11 * 13^55 mod 11 =
3 * 10 mod 11 =
8
however even though it was broken down a bit those are still huge numbers and i done them in a calculator.
I thought of using Fermat's little theorem but the 183 isn't a prime number and from what i understand it needs to be to use that rule?
Any pointers?
I know less than you do, but isn't Fermat's little theorem (what a strange name!) used to determine that 1355 mod 11 = 10?
I think I've learned enough for today.
[Edited by - JohnBolton on June 5, 2006 6:35:42 PM]
1355 mod 11 ( 1311 mod 11 )5 mod 11 <-- ap mod p = a mod p ( 13 mod 11 )5 mod 11 25 mod 11 10 I think I've learned enough for today.
[Edited by - JohnBolton on June 5, 2006 6:35:42 PM]
lets take a smaller example, it follows that your larger numbers work in the same exact way.
5^3 mod 4 = 125 mod 4 = 1
but we can also say 5^3 = 5*5*5, and mod is distributive in a way
so
((((5 mod 4) * 5) mod 4)*5) mod 4) =>
(1 * 5 mod 4) * 5 mod 4 =>
(1) * 5 mod 4 =>
1
You can shorten the process with a square and multiply algorithm to remove some of the multiplications.
[Edited by - KulSeran on June 5, 2006 7:43:11 PM]
5^3 mod 4 = 125 mod 4 = 1
but we can also say 5^3 = 5*5*5, and mod is distributive in a way
so
((((5 mod 4) * 5) mod 4)*5) mod 4) =>
(1 * 5 mod 4) * 5 mod 4 =>
(1) * 5 mod 4 =>
1
You can shorten the process with a square and multiply algorithm to remove some of the multiplications.
int sqpow ( int base, int exp, int n ){ vector<bool> bits; for ( int cnt1 = 0; cnt1 < sizeof(int)*8; cnt1++ ) { bool bit = exp & 1; exp = exp >> 1; bits.push_back(bit); } int rVal = 1; while ( bits.back() == 0 ) bits.pop_back(); int sz = bits.size(); for ( int cnt1 = 0; cnt1 < sz; cnt1++ ) { bool bit = bits.back(); bits.pop_back(); if ( bit ) { rVal *= base; } else { rVal *= 1; } rVal %= n; if ( cnt1 != sz -1 ) { rVal *= rVal; } rVal %= n; } return rVal; }[Edited by - KulSeran on June 5, 2006 7:43:11 PM]
Author
so what about doing 13^(2^7) using KulSeran method then? as its a power to a power?
sorry if this is all daft, it's late as anythin lol and ive been drinkin.
sorry if this is all daft, it's late as anythin lol and ive been drinkin.
Author
read all that while sober and it makes perfect sense lol.
cheers guys, i'll be less stupid in future!
cheers guys, i'll be less stupid in future!
You can't use fermat's little theorem directly, but you can use a result of it, namely that, if gcd(a, c) = 1, then a^b = a^(b mod phi(c)) (mod c). When working with modular equations, you can always replace the base, and any other numbers not rooted within (like exponents, logs, etc) with any number that is congruent mod c. phi() is the Euler function. For instance
13^183 (mod 11)
= 2^183
phi(11) = 10
183 = 3 (mod 10)
2^183 = 2^3 (mod 11)
= 8
In the second example,
5*5*5 (mod 4)
5 = 1 (mod 4)
You can replace all the 5s with 1 since they are congruent with the modulus.
5*5*5 = 1*1*1 = 1 (mod 4)
13^183 (mod 11)
= 2^183
phi(11) = 10
183 = 3 (mod 10)
2^183 = 2^3 (mod 11)
= 8
In the second example,
5*5*5 (mod 4)
5 = 1 (mod 4)
You can replace all the 5s with 1 since they are congruent with the modulus.
5*5*5 = 1*1*1 = 1 (mod 4)
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