To ensure privacy using secret key encryption between N people, each person needs a distinct secret key shared only with one other specific person. This means we need to find the number of unique pairs that can be formed from N individuals.
The problem boils down to selecting groups of 2 people from a total of N people, where the order of selection does not matter. This is a classic combination problem.
The formula for combinations is:
$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $
Here, $n$ represents the total number of people ($N$), and $k$ represents the size of the group for each secret key (which is 2, as a key is shared between a pair).
Substituting $n = N$ and $k = 2$ into the formula:
Number of keys = $ \binom{N}{2} = \frac{N!}{2!(N-2)!} $
Now, we simplify the expression:
Number of keys = $ \frac{N \times (N-1) \times (N-2)!}{2 \times 1 \times (N-2)!} $
Canceling out $(N-2)!$ from the numerator and denominator:
Number of keys = $ \frac{N(N - 1)}{2} $
Therefore, the total number of secret keys required for N people to use secret key encryption for privacy is $ \frac{N(N - 1)}{2} $.
| 1. | DES | a. | 128-256 bits |
| 2. | Rj indeal | b. | 1-448 bits |
| 3. | RC4 | c. | 56 bits |
| 4. | Blowfish | d. | 1-2048 bits |
In the RSA public key cryptosystem, the private and public keys are $(e, n)$ and $(d, n)$ respectively, where $n = p \times q$ and $p$ and $q$ are large primes. Besides, $n$ is public and $p$ and $q$ are private. Let M be an integer such that $0 <M< n$ and $\phi(n) = (p-1)(q - 1)$. now consider the following equations.
I. M’ =\( M^e\) mod n M =\( (M')^d\) mod n
II. ed ≡1 mod n
III. ed ≡1 mod f(n)
IV. M’ = \(M^e \)mod f(n)
M = \( (M')^d\) mod
Which of the above equations correctly represent RSA cryptosystem?