**Challenge Description: I found a public key and ciphertext on the ground near a restaurant in South Boston. Can you decrypt it for me? It shouldn't take too much effort.**
Given ecnryption file , ciphertext, public key files```from Crypto.PublicKey import RSA
def encrypt_flag(plaintext_bytes):    RSAkey = RSA.generate(1024)    n = getattr(RSAkey.key, 'n')    e = 5    plaintext_bytes += b'\x00'    plaintext = int.from_bytes(plaintext_bytes, 'big')    ciphertext = pow(plaintext, e, n)    open('ciphertext', 'w').write(str(ciphertext))    public_key = str(n) + ":" + str(e)    open('public_key', 'w').write(public_key)```Public key is,```n = 105764039675765007162224197946041421610988226034822789741202355465038405474039844301402146302908742536731331641437484787719599778194205333004482617077526529379473501342486898353691458150850096153562792549383987722885036435071184194348535804171098527517150958992100793020614109813938620093243709325590796177891e = 5```CIphertext is,```c = 40030182544273856015788999062464973403472186630147528555052489762516210821795493031619376345647069575950526306492922573846162431037037824967074058132327917359025595463728944947118480605422897682821384491771926743103021286982319660969379132360886299787840185308892024028684314873509707776```
As given challenge name, don't think much to solve the challenge.
**The Vulneralability:**Here, e is very small. So that's the vulneralability.So,plaintext is just the eth root of the ciphertext!Here is complete exploit,```from Crypto.Util.number import *import gmpy
c = 40030182544273856015788999062464973403472186630147528555052489762516210821795493031619376345647069575950526306492922573846162431037037824967074058132327917359025595463728944947118480605422897682821384491771926743103021286982319660969379132360886299787840185308892024028684314873509707776e = 5m = gmpy.root(c,e)[0]flag = long_to_bytes(m)print flag#UMDCTF-{f1x_y0ur_3xp0s}```**Flag:** UMDCTF-{f1x_y0ur_3xp0s}