Time for RSA!

e = 65537

n = 134023913680045880492110426626164090971954352532495944119602241841766743315344885078078359876157853261789964632342961459801834169156073972150056251259429043527344585589350222304649100454018523375146422111308080990153227407607374257909945989989405880451908900962388521742688809203045971595430040363546058882461

c = 28029822339281125656746130462465126562337695724847502110361462137424051877785282190017747622367185811734822848358173889752744532793526865170001730008026265012114272142868291439355242497819138690415609825725896275747305098862360965342890612304634184172461662826694263502110523984165435588581876100119126419239

p = 10982621221489294931830537773519582919197608543135586716536756890800033254598710943732068869355188441831549541090720220093538940987353399632224377192068317

Flag Format: /flag{.+}/

## Solution

RSA is a cryptosystem which allows for secure data transmission. RSA mangles
the `plain text`

in a form that makes it `undecipherable`

to people without the
key, while making it much easier for people with a key to open it and get back
the message.

The mathematics is out of the scope of this article, I would encourage a serious reading on your own to figure it out.

The steps to generate a RSA key is as follows.

- Generate two large prime numbers
`p`

, and`q`

. `n`

=`p`

*`q`

.`e`

is a number co-prime to`lcm(p -1, q - 1)`

and 1 <`e`

<`n`

`d`

=`1 / e mod (p - 1)(q - 1)`

The `1 / e`

in this case, should not be confused with division. It is the
`inverse_modulo`

operation, I.E. `d`

is the number, which when multiplied with
`e`

has the following relation.

```
ed = 1 mod (p - 1)(q - 1)
```

This notation will also be confusing for a newbie to modular arithmatic. What this means is that

```
(e * d) mod (p - 1)(q - 1) = 1
```

The public key consists of `n`

and `e`

. The secret key consists of the rest.
Namely, `d`

, `p`

, `q`

.

The security of RSA relies on the one-way trap function of multiplication.
Which means that for a function `f`

```
f((x, y)) = x * y | x, y ∈ Z
```

It is easy enough to go from `(x, y)`

to `x * y`

(multiplication). But is is
far more difficult to go from `x * y`

to `(x, y)`

(factorization).

The encryption for an message, converted to an integer form, `m`

is given as…

```
c = RSA_e_n(m) = m ^ e mod n
```

`c`

is the cipher text. `RSA_e_n`

is a notation used to denote that the RSA
parameters of `e`

, and `n`

are constant and a feature of RSA instead of being
a part of the arguments to a RSA function.

To get back the `m`

from the `c`

we require the presence of `d`

, which is a good
hint as to why it is a part of the `private`

key in the first place.

```
m' = inverse_RSA_e_n_d(c) = c ^ d mod n
```

It is also now interesting to note why `p`

and `q`

are part of the `private`

key. That comes because if you have either of `p`

or `q`

, generating a `d`

for the RSA key is trivial. Because you can easily find the other factor, and
quickly generate the `d`

via doing a modular_inverse operation of e on base of
(p - 1) * (q - 1).

Once `d`

is obtained, decryption can be done since you now have the key.

Since this is the introductory RSA problem, I am kind enough to give the `p`

directly, which means that this RSA is totally broken.

I also give the following number to integer conversion scheme.

```
A -> 0x41 = 64
AA -> 0x4141 = 16705
AAAAAA -> 0x4141414141 = 280267669825
```

Which basically says that the number is converted to a hexadecimal representation, and then the hexadecimal is intepreted as an ASCII string.

So now, it's time for some `sage`

. Sage Math is a number crunching library with
a ton of useful functions built-in.

There are many alternatives to give, but it is a personal preference of quite a few people to use sage, and it is not without reason.

```
e = 65537
n = 134023913680045880492110426626164090971954352532495944119602241841766743315344885078078359876157853261789964632342961459801834169156073972150056251259429043527344585589350222304649100454018523375146422111308080990153227407607374257909945989989405880451908900962388521742688809203045971595430040363546058882461
c = 28029822339281125656746130462465126562337695724847502110361462137424051877785282190017747622367185811734822848358173889752744532793526865170001730008026265012114272142868291439355242497819138690415609825725896275747305098862360965342890612304634184172461662826694263502110523984165435588581876100119126419239
p = 10982621221489294931830537773519582919197608543135586716536756890800033254598710943732068869355188441831549541090720220093538940987353399632224377192068317
q = n / p
d = inverse_mod(e, (p - 1) * (q - 1))
m = pow(c, d, n)
print(m)
```

This prints out `240545625414703578862070172273428889513126431163886829837844391499244541180606084628879027055096318636117555581`

.

Converting this to integer is done by.

```
print(bytes.fromhex(hex(240545625414703578862070172273428889513126431163886829837844391499244541180606084628879027055096318636117555581)[2:]))
```

Which prints out `b'flag{breaking_rsa_is_easy_if_you_know_the_key}'`

. Which is the flag.

## Flag

flag{breaking_rsa_is_easy_if_you_know_the_key}

Recommended Reading

# Foren-Steg

*#ctf #forensics #stegno #zip*

Find the flag, flag finder.

- Flag Format: /flag{.+}/

Provided foren-steg.docx

## Solution

The file provided to us appears to be a docx file. Let us see what happens when we try to open it.

There appears to be nothing in the file but gibberish...

...# Breaking Random Number Generators with Chosen Seed

*#crypto #ctf #misc #python*

Find the flag.

Source is as follows

```
#!/usr/bin/env python3
import random
import time
import string
import signal
# use secure seed
random.seed(int(time.time()))
with open('flag.txt') as f:
flag = f.read()
# large constant prime
p = 174807157365465092731323561678522236549173502913317875393564963123330281052524687450754910240009920154525635325209526987433833785499384204819179549544106498491589834195860008906875039418684191252537604123129659746721614402346449135195832955793815709136053198207712511838753919608894095907732099313139446299843
```