- 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 = 2953input = [14, 17, 20, 23, 14, 11, 20, 23, 14, 26, 11]
seckey = 3041input = [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}