- Plain text (input) is a list of eleven numbers.
- Both secret key and public key are prime numbers from 2806 to 3060.
- Public key is 2909. The secret key is other than 2909.
- The encryption algorithm is as follows.

CipherText = (seckey x 11 x input) % pubkey + randint(1 to 3)

---
I programed a simple Python script and got the following two answers.

```
seckey = 2953
input = [14, 17, 20, 23, 14, 11, 20, 23, 14, 26, 11]

seckey = 3041
input = [4, 5, 6, 7, 4, 3, 6, 7, 4, 8, 3]
```
But neither answer was wrong.
At this time, I did not know the reason.

---
When I checked the problem around 7:00[UTC], the problem sentence was changed.

Before:

```
The flag will be 'flag{sha256 of the concatenated digits of the plain text}'.

Cipher: 1445, 2897, 1438, 2890, 1444, 2901, 1439, 2891, 1444, 1435, 2902

Public Key: 2909
```

After:

```
Cipher Text: [483, 1144, 1803, 1323, 483, 2643, 1802, 1983, 484, 1804, 2641]

Public Key: 2819

The flag will be 'flag{md5 of the secret key}'.
```

The correct answer was below.

flag{md5(2879)}

= flag{ec8b57b0be908301f5748fb04b0714c7}