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How to get mod of large numbers?

Asked by Mohsin Shah on 1 May 2017
Latest activity Commented on by Mohsin Shah on 2 May 2017
I have this calculation: c=(5652)^42.(23)^77mod5929, when i use scientific calculator I get c=4624 an when I do this by MATLAB mod command it gives 0...why?

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2 Answers

Answer by Steven Lord
on 1 May 2017
 Accepted Answer

If you compute (5652^42)*(23^77) to use the naive approach for computing this quantity, the resulting number is much greater than flintmax. If we look at the spacing between that number and the next largest number, we see that the spacing is (much) greater than 5929.
>> x = (5652^42)*(23^77)
x =
2.78954779454946e+262
>> eps(x)
ans =
3.49595995098571e+246
>> (x + 5929) == x
ans =
logical
1
As Sean stated in this answer from a while ago, you could use Symbolic Math Toolbox. Or you could use one of the algorithms described in this Wikipedia page to perform these computations with numbers no greater than 5652^2.

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Thank you so much

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Answer by John D'Errico
on 1 May 2017
Edited by John D'Errico
on 1 May 2017

Because MATLAB works in double precision.
5652^42 * 23^77
ans =
2.78954779454946e+262
This is a number with 263 decimal digits, so far beyond the precision available to a double. A double can represent no integer exactly that is larger than 2^53. And 2^53 is just not very large in this context.
You can either use a tool (like my VPI toolbox or the symbolic toolbox)
vpi(5652)^42 * vpi(23)^77
ans =
27895477945494593633174830929266408591443695665432884022901567138564930227891230414671987539540619495363075575420684353178819655415398533495806892010942366092828416772086484568033389306894581602342651519614654523921262520198453838046500944165540916333162855923712
mod(ans,5929)
ans =
4624
Or, you can recognize that you can use mod in a loop. Thus, this law applies:
mod(a*b,n) = mod(mod(a,n)*mod(b,n),n)
So you can compute the mod of a power (or a product of powers) in a simple loop, while never exceeding the limits imposed by double precision.

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Thank you so much

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