how to calculate modulus in matlab ?

Hi everyone,

the question seems simple as using mod works for small numbers not for large numbers.

i want to calculate mod( (4^15)*(21^13),47) the matlab ans= 21 but the correct ans = 3 using the windows calculator. is there any way in matlab to calculate such modulus ?

Thanks in advance.

Answer 1

[James Tursa]

See this newsgroup thread where Bruno Luong gives advice on how to do this calculation without using the symbolic toolbox:

http://www.mathworks.com/matlabcentral/newsreader/view_thread/329023

Answer 2

[Roger Stafford]

The number you describe is far too large for accurate numerical computation using only 'double' floating points numbers. However, the 'mod' function also works with symbolic numbers using the symbolic toolbox. You can compute with these to any accuracy you wish.

Answer 3

[Walter Roberson]

mod(mod( (4^15), 47) * mod((21^13),47), 47)

See also

http://www.mathworks.com/matlabcentral/newsreader/view_thread/253630

http://www.mathworks.com/matlabcentral/fileexchange/7908-big-modulo-function

http://www.mathworks.com/help/symbolic/mupad_ref/powermod.html

http://www.mathworks.com/matlabcentral/fileexchange/39716-powermod

http://www.mathworks.com/matlabcentral/fileexchange/38516-powermod

[Roger Stafford]

The number 21^13 as computed with 'double' is too large to be computed exactly. Its least five bits must necessarily be rounded to zeros to fit within 53 bits and consequently it gives:

 mod(21^13,47) = 2,

whereas the true modulo 47 value is:

 mod(mod(21^7,47)*mod(21^6,47),47) = 7.

Answer 4

[ahmed]

Thanks for your help any way , but even the powermod function or any of the above functions in url links can calculate the correct answer which is 3 .

trying:

powermod( (4^15) * (21^13) , 1 , 47 ) gives 21 but the correct ans using windows calculator is 3 and it is equivelant to the answer in my cryptography study book .

[Walter Roberson]

mod(powermod(4, 15, 47) * powermod(21, 13, 47), 47)

ans = 3