crypto/4096
360 Points
I heard 4096 bit RSA is secure, so I encrypted the flag with it.
Taking a look source.py and output.txt, we know that
n = 50630448182626893495464810670525602771527685838257...
pow(m, e, n) = 156406298972120895391457696256321891254...
and note that n is the product of 128 primes, all of which are relatively small (2^32). Thus, use https://www.alpertron.com.ar/ECM.HTM to find the 128 prime factors of note. Note that this may have to be used multiple times, as it soemtimes gets "stuck" on medium sized primes.
Now the list of primes is in solution.py, along with n, e, c. And using https://gist.github.com/jackz314/09cf253d3451f169c2dbb6bbfed73782 it can be found that.
m = 54372504422578163821842661820992519574720094743792971271580036722643401901785509893350887232430564989
thus doing Decimal -> Hex with https://www.rapidtables.com/convert/number/decimal-to-hex.html
m = 636F726374667B746F305F6D346E795F7072316D337335355F363361656561333761366233623232667D
and doing Hex -> ASCII with https://www.rapidtables.com/convert/number/hex-to-ascii.html
This gives the desired flag,
corctf{to0_m4ny_pr1m3s55_63aeea37a6b3b22f}