Strong Encryption algorithm

[jon_davis24]

I was watching a show on TLC the other day about encryption and cryptography. It discussed the problems of distributing keys for un-encryption. It also showed an encryption algorithm developed that is (if I remember correctly) similar to the RSA method. The formula is:

m13 (mod k) = e

m = the message in numbers (i.e. x = 88)
Note: the m13 is the message (m) to the power of 13 (m * m * m * m * m * m * m * m * etc . . .) but I don't have superscript on my keyboard.

k = the key (any number . . . i.e. 0.0000178578 to 586839588939 or longer)

e = the encrypted message

It is pretty simple . . . the message (m) is multiplied by itself 13 times, then the resulting number is multiplied by the key, which is first inserted in the (mod ). The result is extremely difficult to decrypt without the key.

Now, my question is, how would I use (mod ) in Visual Basic?

Thanks,

JD

[unformed]

well if you're trying to do:
n (mod k)

the code would be
n Mod k

pretty straightforward...

[kedaman]

the problem is that a message^13 is too big for vb to handle if it exceeds 4 bits.

[unformed]

oh yeah, guess i should have read the whole question ;) ...

there's a trick for this, i'll have to look it up...but i'll get back to you soon

btw...if you're into cryptography, read Bruce Schnier's (sp?) "Applied Cryptography" ... amazing book...

[nishantp]

your probably better of in C++ cause you can write code to support those huge nums.

[eiSecure]

by the way, it's spelled Bruce Schneier. :)