# Cryptography

• May 17th 2008, 08:07 AM
flinted
Cryptography
I was also wondering if someone could help me find the answers to these simple problems

$\displaystyle -16 mod 3$

$\displaystyle -52 mod 19$

is there a formula to follow for these problems?

Could someone help me get a solution to this probelm.

Compute $\displaystyle 2^{1000000} mod 7$

Repeated Square method might do it.
• May 17th 2008, 08:10 AM
Moo
Hello,

Quote:

Originally Posted by flinted
Could someone help me get a solution to this probelm.

Compute $\displaystyle 2(^1000000) mod 7$

$\displaystyle 2^3=8=1 \mod 7$

1000000=3k+1

Therefore, $\displaystyle 2^{1000000}=2^{3k+1}=2 \cdot (2^3)^k \equiv 2 \cdot 1^k \mod 7$

$\displaystyle \boxed{2^{1000000} = 2 \mod 7}$
• May 17th 2008, 08:12 AM
Isomorphism
Quote:

Originally Posted by flinted
Could someone help me get a solution to this probelm.

Compute $\displaystyle 2(^)((1000000)) mod 7$

To find $\displaystyle 2^{1000000} mod 7$, observe that $\displaystyle 2^3 = 1 mod 7$. This means $\displaystyle 2^{999999} = (2^3)^{333333} = 1^{333333} mod 7$.$\displaystyle 2^{999999} = 1 mod 7 \Rightarrow 2^{1000000} = 2 mod 7$
• May 17th 2008, 08:17 AM
flinted
thanks but i dont undertand where you got the $\displaystyle k + 1$ formula from
• May 17th 2008, 08:21 AM
Moo
1000000=3x333333+1
I substitute 333333 by k.
• May 17th 2008, 08:26 AM
flinted
thanks alot
• May 19th 2008, 07:16 AM
flinted
*bump*