Rating:

This looks like some RSA with "small numbers" ;)

We can easily factorise n. Once we know n we can compute phi.

```

n=960242069=151*6359219

Phi = (p-1)(q-1) = 150*6359218 = 953882700

e = 347

ed = 1 mod 953882700

```

Then, as we already know e we can compute d with a simple python loop.

```

>>> i = 347

>>> while (347*i%953882700 !=1):

... i+=1

...

>>> i

5497883

```

We create a list c of the number as in the challenge description and decode every element using our RSA numbers.

```

>>> c=[346046109,295161774,616062960,<SNIP>,770057231,770121847]

>>> for elem in c:

... print(chr((elem**5497883)%960242069))

```

This was a bit long as a single process is involved but at the end we got the following output: xhBQCUIcbPf7IN88AT9FDFsqEOOjNM8uxsFrEJZRRifKB1E=|key=visionary

Vigenere cipher using the key visionary.

Once decrypted with the key we get the following: zJIOHIldUx7QF88MG9FMHxiMGAwNV8wckNjQWZATnxST1Q=

We decrypt the base64:

```

$ echo -ne 'czJIOHIldUx7QF88MG9FMHxiMGAwNV8wckNjQWZATnxST1Q=' | base64 -d

s2H8r%uL{@_<0oE0|b0`05_0rCcAf@N|ROT

```

And then a ROT47 give us the flag: DawgCTF{Lo0k_@t_M3_1_d0_Cr4p7o}

Original writeup (https://maggick.fr/2020/04/dawgctf-2020.html).