Tags: cryptography-rsa
Rating:
# Sophie Wilson
## Description
> Sophie Mary Wilson CBE FRS FREng DistFBCS (born Roger Wilson; June 1957) is an English computer scientist, who helped design the BBC Micro and ARM architecture. Wilson first designed a microcomputer during a break from studies at Selwyn College, Cambridge. She subsequently joined Acorn Computers and was instrumental in designing the BBC Micro, including the BBC BASIC programming language whose development she led for the next 15 years. She first began designing the ARM reduced instruction set computer (RISC) in 1983, which entered production two years later. - Wikipedia Entry
> Chal: Help this designer of microprocessors solve this RSA challenge.
```python
n = 784605825796844081743664431959835176263022075947576226438671818152943359270141637991489766023643446015742865872000712625430019936454136740701797771130286509865524144933694390307166660453460378136369217557779691427646557961148142476343174636983719280360074558519378409301540506901821748421856695675459425181027041415137193539255615283103443383731129040200129789041119575028910307276622636732661309395711116526188754319667121446052611898829881012810646321599196591757220306998192832374480348722019767057745155849389438587835412231637677550414009243002286940429895577714131959738234773350507989760061442329017775745849359050846635004038440930201719911010249665164009994722320760601629833907039218711773510746120996003955187137814259297909342016383387070174719845935624155702812544944516684331238915119709331429477385582329907357570479058128093340104405708989234237510349688389032334786183065686034574477807623401744101315114981390853183569062407956733111357740976841307293694669943756094245305426874297375074750689836099469106599572126616892447581026611947596122433260841436234316820067372162711310636028751984204768054655406327047223250327323182558843986421816373935439976256688835521454318161553726050385094844798296897844392636332777
e = 5
c = 268593521627440355433888284074970889184087304017829415653214811933857946727694253029979429970950656279149253529187901591829277689165827531120813402199222392031974802458605195286640398523506218117737453271031755512785665400604866722911900724895012035864819085755503886111445816515363877649988898269507252859237015154889693222457900543963979126889264480746852695168237115525211083264827612117674145414459016059712297731655462334276493
```
## Solution
* We are given with the following RSA parameters:
* `n`: Modulus
* `e`: Public Exponent
* `c`: Cipher text
* This is vulnerable to `small e attack`.
* As the public exponent is small, taking the **eth** root of the ciphertext gives back the original plaintext message `m`.
* When the public exponent is small, we dont need the private key for decryption.
* We can use the following python script to get the flag:
[script.py](https://github.com/HarshJolad/CTF-Writeups/blob/master/CyberHeroines-CTF-2023/crypto/Sophie_Wilson/script.py)
```python
from Crypto.Util.number import long_to_bytes
from gmpy2 import iroot
e = 5
c = 268593521627440355433888284074970889184087304017829415653214811933857946727694253029979429970950656279149253529187901591829277689165827531120813402199222392031974802458605195286640398523506218117737453271031755512785665400604866722911900724895012035864819085755503886111445816515363877649988898269507252859237015154889693222457900543963979126889264480746852695168237115525211083264827612117674145414459016059712297731655462334276493
# Calculate the fifth root of c
ith_root = iroot(c, e)[0]
# Convert to a hexadecimal string and then decode it
pt = bytes.fromhex(hex(ith_root)[2:]).decode()
print("Plaintext: ", pt)
```
### FLAG
```
chctf{d3516n3d_4c0rn_m1cr0_c0mpu73r}
```
