# Quick Math

by anishbadhri

Ben has encrypted a message with the same value of ‘e’ for 3 public moduli

- n1 = 86812553978993
- n2 = 81744303091421
- n3 = 83695120256591

and got the cipher texts

- c1 = 8875674977048
- c2 = 70744354709710
- c3 = 29146719498409

Find the original message. (Wrap it with csictf{})

## Solution

The given problem is a typical example of hastad attack. Given 3 pairs of `n`

and `c`

with the same value of `e=3`

, the original message can be decoded.

**Solution Script**:

```
from pwn import remote
from sympy.ntheory.modular import crt
from gmpy2 import iroot
e = 3
N = [86812553978993, 81744303091421, 83695120256591]
C = [8875674977048, 70744354709710, 29146719498409]
resultant, mod = crt(N,C)
value, is_perfect = iroot(resultant,e)
print(bytes.fromhex(str(value)).decode())
```

## Flag

```
csictf{h45t4d}
```

*Crypto*