# TSG CTF 2020 Beginner's Crypto
## Problem
The problem is given in the form of the follewing file:
```pythonassert(len(open('flag.txt', 'rb').read()) <= 50)assert(str(int.from_bytes(open('flag.txt', 'rb').read(), byteorder='big') << 10000).endswith('1002773875431658367671665822006771085816631054109509173556585546508965236428620487083647585179992085437922318783218149808537210712780660412301729655917441546549321914516504576'))```
The flag is converted to an integer value and shifted 10,000 bits to the left to give the last 175 decimal digits of the value. In other words, the purpose of this problem is to restore the flag value from:
$$c=\text{flag}\cdot2^{10000}\bmod10^{175}$$
## Solution
At first you may try taking the inverse of $2^{10000}$ over $F_{10^{175}}$, but that should be impossible because they are not coprime.
The correct way is, first think of
$$c=\text{flag}\cdot2^{10000}\bmod5^{175}$$
and, using Euler’s theorem, calculate:
$$\begin{aligned}(2^{10000})^{-1}\bmod5^{175}&=(2^{10000})^{\varphi(5^{175})-1}\\&=(2^{10000})^{5^{175}-5^{174}-1}\end{aligned}$$
and get the flag by:
$$\text{flag}=c\cdot(2^{10000})^{-1}\bmod5^{175}$$
Since ${256}^{50}<5^{175}$, this is the only value that can satisfy the equation.
### Solver Script
```pythonc = 1002773875431658367671665822006771085816631054109509173556585546508965236428620487083647585179992085437922318783218149808537210712780660412301729655917441546549321914516504576mod = 5 ** 175phi = 5 ** 175 - 5 ** 174inv = pow(pow(2, 10000, mod), phi - 1, mod)print(((c * inv) % mod).to_bytes(50, byteorder='big'))```
The answer is `TSGCTF{0K4y_Y0U_are_r3aDy_t0_Go_aNd_dO_M0r3_CryPt}`.
## Alternative Solution #1 (by [@naan112358](https://twitter.com/naan112358))
Let's start with
$$c=\text{flag}\cdot2^{10000}\bmod5^{175}$$
for the previous solution.
By considering:
$$2x=\begin{cases}2x\qquad\ \ (2x<M)\\2x-M\ (2x\geq M)\end{cases}\bmod M$$
we can assume that the above case produces even number and the below case produces odd number. So,
```pythondef divide_by_two(x):    if x % 2 == 0:        return x // 2    else:        return (x + mod) // 2```
simulates division by 2 over the given modulo.
We can repeat this 10000 times and get the flag.
### Solver Script
```pythonmod = 5 ** 175c = 1002773875431658367671665822006771085816631054109509173556585546508965236428620487083647585179992085437922318783218149808537210712780660412301729655917441546549321914516504576 % mod
def divide_by_two(x):    if x % 2 == 0:        return x // 2    else:        return (x + mod) // 2
for i in range(10000):    c = divide_by_two(c)
print(c.to_bytes(50, byteorder='big'))```
## Alternative Solution #2 (by [@satos___jp](https://twitter.com/satos___jp))
Now we put:
$$C=\text{flag}\cdot2^{10000}$$
Note that there's no modulo here.
On the other hand we know $c$, which is the last 175 digits of $C$, so we can put:
$$C=c_0\cdot{10}^{175}+c$$
Now we want to calculate $c_0$ which makes the last 10000 digits of the binary form of $C$ all zero.
Then consider $C'$, which can be obtained by replacing $c_0$ to $c'_0=c_0+2^x$. Now we get:
$$\begin{aligned}C'&=c'_{0}\cdot10^{175}+c\\&=\left(c_{0}+2^{x}\right)\cdot10^{175}+c\\&=c_{0}\cdot10^{175}+c+2^{x}\cdot10^{175}\\&=C+5^{175}\cdot2^{x+175}\end{aligned}$$
And the last $x+175$ digits of the binary form of $C$ is unchanged.
So, by determining $c_0$ from the least significant bits, we can obtain $C$ such that all the last 10000bit is zero.
Flag can be obtained by right-shifting $C$ 10000bits.
### Solver Script
```pythonc = 1002773875431658367671665822006771085816631054109509173556585546508965236428620487083647585179992085437922318783218149808537210712780660412301729655917441546549321914516504576c0 = 0bit = 1mask = 1 << 175while '1' in bin(c0 * 10 ** 175 + c)[-10000:]:  if (c0 * 10 ** 175 + c) & mask != 0:    c0 += bit  bit <<= 1  mask <<= 1C = c0 * 10 ** 175 + cflag = (C >> 10000).to_bytes(50, byteorder='big')print(flag)```