ElGamal Example (explained easier)

This scheme requires several variables.

a is a secret key less than the value of p. We will assume it is 304 since p must be greater than it.
g is the generator variable. Nokia has told us its value is 10.
p is a large prime number that Nokia have told us is >304.
For our situation, p is the most important. It is a prime number that must have 'g' as a primitive root.
For fun, we will select 2789.
m is our encrypted message (if there is one) that has to be less than p. For fun we'll choose 832 (from taxi)
k is a prime number used for encrypting the message. It also has the condition it must be less than p and gcd(k, p − 1) = 1. For fun, we will choose 15 (from truck).

So let's do it.

Generate Public Key

p = 2789; g = 10; a = 304
g^a % p = 10^304 (% 2789) ≡ 1161

Now we have what is known as a 'public key' (p, g, g^a%p).
The public key is P = (p ← 2789, g ← 10, g^a%p ← 1161), the secret key is a (304).

Encrypt Message

Next, we encrypt a message m = 832 as follows:
k = 15
γ = 10^15 (% 2789) ≡ 1063
δ ≡ 832 x 1161^15 (% 2789) ≡ 181
c = (γ ← 1063, δ ← 181)

The ciphertext, c, is then sent to the recipient.

Decrypt Message

The recipient can then calculate m from this. The message is calculated by:
m ≡ 1063^(2789-1-304) x 181 ≡ (1063^2484) x 181 (% 2789) ≡ 832

In summary -- to get your 6 digits for the competition:
Generating Public Key
<p><10><10^a % p> (6 digits)

In this situation, p is most likely 3 digits. Also, a or y is likely 15 (where else would it be used?)
This means you can get p from the list below:
313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619, 647, 659, 701, 709, 727, 743, 811, 821, 823, 857, 863, 887, 937, 941, 953, 971, 977, 983.

Generating Cipher Key
<10^k % p><m*(10^a%p)^k %p> (6 digits)

In this situation, p is most likely 4 digits.
This means Nokia has given us the p from the 'colors on the wall'.