# TJCTF 2019 "All the Zips" writeup
## check problem
```140 zips in the zip, all protected by a dictionary word.
All the zips```
I get 140 zip file.
## solve problem
It is said that password is a dictionary word.
I try to crack password by fcrackzip.
```$ fcrackzip -u -D -p /usr/share/dict/words zip0.zipPASSWORD FOUND!!!!: pw == how```
```$ unzip -o -P "how" zip0.zip ; cat flag.txtArchive:  zip0.zipextracting: flag.txtwhew```
I can easily get password, and unzip file.
Then, let's try all 140 zip files!
```bash#!/bin/bash
for i in {0..139} ; do
  fcrackzip -u -D -p ./words zip$i.zip >> pass.log
  echo "zip$i.zip" >> pass.log
done```
```bash#!/bin/bash
unzip -o -P "how" zip0.zip ; cat flag.txt >> flag.log ; echo "" >> flag.log
unzip -o -P "to" zip1.zip ; cat flag.txt >> flag.log ; echo "" >> flag.log
unzip -o -P "establish" zip2.zip ; cat flag.txt >> flag.log ; echo "" >> flag.log
・・・```
I can get flag in zip85.zip.
```$ unzip -o -P "every" zip85.zip ; cat flag.txttjctf{sl4m_1_d0wn_s0_that_it5_heard}```