# picoCTF Mind your Ps and Qs Write Up
## Details:Points: 20
Jeopardy style CTF
Category: Cryptography
Comments: In RSA, a small e value can be problematic, but what about N? Can you decrypt this?
## Write up:
Using cat on the file I got:
```cat values           
Decrypt my super sick RSA:c: 62324783949134119159408816513334912534343517300880137691662780895409992760262021n: 1280678415822214057864524798453297819181910621573945477544758171055968245116423923e: 65537  ```
This was a pretty small n value so I decided to use alpertron to find the prime factors:
1899107986527483535344517113948531328331 and 674357869540600933870145899564746495319033
I then plugged these values and the c, n, and e into the rsa cracker script I wrote for dvCTF:
```python# all prime factorsx = [1899107986527483535344517113948531328331, 674357869540600933870145899564746495319033]
# nn = 1280678415822214057864524798453297819181910621573945477544758171055968245116423923
# ee = 65537
# cipher textct = 62324783949134119159408816513334912534343517300880137691662780895409992760262021
# start value for phi(n)i = 1
# loop through prime factors and multiply them together with (factor-1)*(nextFactor-1)...for a in x:	i = i * (a-1)
# inverse pow (3.8+ syntax, for previous versions of python use gmpy2.invert)d = pow(e, -1, i)
# solve for the answerans = pow(ct, d, n)
# print the answerprint(bytes.fromhex(hex(ans)[2:]).decode('ascii'))```
When run I got:
```picoCTF{sma11_N_n0_g0od_05012767}```