This looks like some RSA with "small numbers" ;)
We can easily factorise n. Once we know n we can compute phi.
```n=960242069=151*6359219Phi = (p-1)(q-1) = 150*6359218 = 953882700e = 347ed = 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...>>> i5497883```
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 -ds2H8r%uL{@_<0oE0|b0`05_0rCcAf@N|ROT```
And then a ROT47 give us the flag: DawgCTF{Lo0k_@t_M3_1_d0_Cr4p7o}
