Rating:
# RSA-2
## Description
```
Hey I heard you have a supercomputer at home. This is taking too long to compute on my computer.
Could you take a look on yours? I’m sure its a lot more precise than mine is, and faster too!
```
supercomputer_food:
```
e: 3
c: 2780321436921227845269766067805604547641764672251687438825498122989499386967784164108893743279610287605669769995594639683212592165536863280639528420328182048065518360606262307313806591343147104009274770408926901136562839153074067955850912830877064811031354484452546219065027914838811744269912371819665118277221
n: 23571113171923293137414347535961677173798389971011031071091131271311371391491511571631671731791811911931971992112232272292332392412512572632692712772812832933073113133173313373473493533593673733793833893974014094194214314334394434494574614634674794874914995035095215235415475575635695715775875935996016076136176196316416436476536596616736776836917017097197277337397437517577617697737877978098118218238278298398538578598638778818838879079119199299379419479539679719779839919971009101310191431936117404941729571877755575331917062752829306305198341421305376800954281557410379953262534149212590443063350628712530148541217933209759909975139820841212346188350112608680453894647472456216566674289561525527394398888860917887112180144144965154878409149321280697460295807024856510864232914981820173542223592901476958693572703687098161888680486757805443187028074386001621827485207065876653623459779938558845775617779542038109532989486603799040658192890612331485359615639748042902366550066934348195272617921683
```
So now encryption using RSA can be mathematically expressed as:
c = m^emod(n)
where mod is the modulus operator
and decryption from RSA can be expressed as:
m = c^d mod(n)
where again mod is the modulus operator
We can see the value of m^e is smaller than the value of n or n is very large than the exponent e.
Therefore the cipher text would be equal to m^e and the effect of encryption is equal to raising the plaintext message to the power of the exponent e.
So we took cuberoot of c using the following python script
```
from decimal import *
n = 23571113171923293137414347535961677173798389971011031071091131271311371391491511571631671731791811911931971992112232272292332392412512572632692712772812832933073113133173313373473493533593673733793833893974014094194214314334394434494574614634674794874914995035095215235415475575635695715775875935996016076136176196316416436476536596616736776836917017097197277337397437517577617697737877978098118218238278298398538578598638778818838879079119199299379419479539679719779839919971009101310191431936117404941729571877755575331917062752829306305198341421305376800954281557410379953262534149212590443063350628712530148541217933209759909975139820841212346188350112608680453894647472456216566674289561525527394398888860917887112180144144965154878409149321280697460295807024856510864232914981820173542223592901476958693572703687098161888680486757805443187028074386001621827485207065876653623459779938558845775617779542038109532989486603799040658192890612331485359615639748042902366550066934348195272617921683
c = 2780321436921227845269766067805604547641764672251687438825498122989499386967784164108893743279610287605669769995594639683212592165536863280639528420328182048065518360606262307313806591343147104009274770408926901136562839153074067955850912830877064811031354484452546219065027914838811744269912371819665118277221
e = 3
c = Decimal(c)
e = Decimal(e)
getcontext().prec = 1024 # Set a big enough precision
root = pow(c, 1/e) # Calculate c^(1/e) = m^(e * 1/e) = m
print(root)
# Decode with no padding
m = hex(int(root))[2:-1] # Number to hex
m = ''.join(chr(int(m[i:i+2], 16)) for i in range(0, len(m), 2)) # Hex to Ascii
print(m)
```
## Flag
```
dsc{t0-m355-w1th-m4th-t4k35-4-l0t-0f-sp1n3}
```
