# b00tl3gRSA2

## Problem

> In RSA d is alot bigger than e, why dont we use d to encrypt instead of e? Connect with nc 2019shell1.picoctf.com 25894.

## Solution

1. Run `nc 2019shell1.picoctf.com 25894`

```
c: 57815126326093923236185780828975923133200522878557452390870456883515440528642167536803747941322559284194471301880831184375049408521327486560563277740668565677490872095919303574108987650457633864377449098953674268225360856713612093782972460048667200729242166386409098664867986711862098908584062960036249053871
n: 157847463302497252662544277589791143185923975522921166059858180482551957816021742934853967462755774626496767368758467309346738171067691945316670158280275671837492151727611658262787392308543243056437019027884283665423329385531974860426338707097143562911124606008275824897108139807573312761136578124929695271821
e: 102798229720827704058624171259134165163471339843047412737855058992260861979425729133199600001188324409623701543646797125749242899115616520714417794918297219527473616153153909309469518422043092916690361135134781912570527561162416572764140527303057017712606139457097065518745602444936727612425446027579066085729
```

1. Use [RsaCtfTool](https://github.com/Ganapati/RsaCtfTool) to run `python RsaCtfTool.py --uncipher 57815126326093923236185780828975923133200522878557452390870456883515440528642167536803747941322559284194471301880831184375049408521327486560563277740668565677490872095919303574108987650457633864377449098953674268225360856713612093782972460048667200729242166386409098664867986711862098908584062960036249053871 -n 157847463302497252662544277589791143185923975522921166059858180482551957816021742934853967462755774626496767368758467309346738171067691945316670158280275671837492151727611658262787392308543243056437019027884283665423329385531974860426338707097143562911124606008275824897108139807573312761136578124929695271821 -e 102798229720827704058624171259134165163471339843047412737855058992260861979425729133199600001188324409623701543646797125749242899115616520714417794918297219527473616153153909309469518422043092916690361135134781912570527561162416572764140527303057017712606139457097065518745602444936727612425446027579066085729`
2. Within a few seconds you should get an output of a whole bunch of zeros followed by the flag. 

### Flag

`picoCTF{bad_1d3a5_3468581}`


---

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