Mod function in a claculator

jank

As the title says is there an easier way to find out the mod of a number using a claculator rahter than then a long winded way!?


Me own claculator is a sharp el-53ILH if its any help at all

jank

anybody

have a cryptography exam tommorrow and i need it so save time

seamus

...

Let's say for example...

You have 133
And you want to find 133 % 8

Divide 133 by 8 = 16.625

Which means it goes evenly in 16 times.

So multiply 16 x 8 and subtract it from 133. Ans: 5

Simple

or...

Back to 16.625.
Take the non-integer part, i.e. 0.625, and multiply it by 8, which will give you 5.

The first one is better though, as the second one may not produce an integer due to rounding.

jank

yea i know that is a way of doing
cheers

but in my exam im going to have to deal with numbers like 2^8
or so
eg. 2^7mod11 !

i know its do able but very long winded:dunno:
was looking for a short-cut

thanks anyway though;)

ecksor

Do you know your rules for modulo arithmetic? As in, [a]n + n = [a+b]n and [a]n * n = [a*b]n

Hard to give you an answer without knowing your course though and your example looks a bit bizarre (well, too small basically). Do you remember drawing cayley tables of groups or anything like that? 2 generates the group like so:

2->4->8->5->10->9->7->3->6->1

If you were given 2^270 mod 11 then you'd look for 270 mod 11 which is 270-(11*24) = 6

So, 2^270 mod 11 = 2^6 mod 11 and according to the cycle above that's 9.

seamus

Originally posted by ecksor
Do you know your rules for modulo arithmetic? As in, [a]n + n = [a+b]n and [a]n * n = [a*b]n

Hard to give you an answer without knowing your course though and your example looks a bit bizarre (well, too small basically). Do you remember drawing cayley tables of groups or anything like that? 2 generates the group like so:

2->4->8->5->10->9->7->3->6->1

If you were given 2^270 mod 11 then you'd look for 270 mod 11 which is 270-(11*24) = 6

So, 2^270 mod 11 = 2^6 mod 11 and according to the cycle above that's 9.

Funny....how did a university give me a Comp Sci degree without ever touching on modulo arithmetic?

ecksor

You'd have to have seen it in something like a crypto module I guess or something else that covers group theory or number theory I suppose ...

Of course, numeric datatypes act like modulo arithmetic too which is where integer overflows and similar bugs come from.

ecksor, degreeless.

jank

Mod is used in RSA public key cryptography.

P^e mod n =C

C^d mod n= p


the prime factors is how one would derive the encryption key from the decryption key ie. get the inverse of it
and visa versa but the strength of the key lies in the fact that RSA standard keys are 1024 bytes long and finding 2 prime factors of a number like that is impossible (well so ive been told until quantum computers are invented)

e*d mod (p-1)(q-1) =1
d=e^-1 mod (p-1)(q-1)


^
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me thinks im smart cause i just learned it
wish me luck tomorrow

ruchika

how do I find mod of a matrice ??