3svgl/rhzp()gz12kmuuzs5fyeseeneymi.

[Epicman9001]

Just nuke the world right now you pussy

[gold stinger]

No I think I want to monologue & bad talk people some more while I have the time we all agreed upon to do it.

[gold stinger]

24 hours to go gentlemen. Will you make it in time? Will you lose a chunk of the planet?

So devilishly close, so very very close...

Check back on hints, clues that people have posted, findings, and answers. Use them wisely



My hint for the last 24 hours:
I didn't use the regular alphabet order.

[reuben_tate]

My keyboard is right in front of me

I got the first word I think with this.

Very promising, gonna see if the rest makes sense with this.

q 0
w 1
e 2
r 3
t 4
y 5
u 6
i 7
o 8
p 9
a 10
s 11
d 12
f 13
g 14
h 15
j 16
k 17
l 18
z 19
x 20
c 21
v 22
b 23
n 24
m 25

[reuben_tate]

if anyone can crack this, I will pay you one hundred thousand credits

payment please

[danceflashrevo]

gmferng

[gold stinger]

Originally Posted by reuben_tate

if anyone can crack this, I will pay you one hundred thousand credits

payment please

That is the correct answer to the code.

[Charu]

NERDS!!!

[gold stinger]

I have sent 200,000 credits to the account of reuben_tate, for completing the code. Feel free to share your winnings as you see fit.

Currently sending something else for reuben in the form of PM. Will post answer to the code for everyone shortly.

[Charu]

Yo, can I just say not all keyboards use QWERTY

wow wow wow wow wow

[DossarLX ODI]

reuben you got the answer a few minutes before I did, dammit haha

For anyone that's interested in what the algorithm was, the shift was +14 except for every third letter. So every third letter was a +13 shift.

# stupid python 3 program for keyboard caesar shift

# Letter to number
qwerty = {
"q": 0,
"w": 1,
"e": 2,
"r": 3,
"t": 4,
"y": 5,
"u": 6,
"i": 7,
"o": 8,
"p": 9,
"a": 10,
"s": 11,
"d": 12,
"f": 13,
"g": 14,
"h": 15,
"j": 16,
"k": 17,
"l": 18,
"z": 19,
"x": 20,
"c": 21,
"v": 22,
"b": 23,
"n": 24,
"m": 25,
}

# Number to letter
numqwerty = {
0:"q",
1:"w",
2:"e",
3:"r",
4:"t",
5:"y",
6:"u",
7:"i",
8:"o",
9:"p",
10:"a",
11:"s",
12:"d",
13:"f",
14:"g",
15:"h",
16:"j",
17:"k",
18:"l",
19:"z",
20:"x",
21:"c",
22:"v",
23:"b",
24:"n",
25:"m",
}

stringToTest = input("Insert goldstinger string: ")
result = ""
counter = 0
for letter in stringToTest:
    if letter == " ":
        result += " "
        continue
    if counter == 2:
        shift = 13
        counter = 0
    else:
        shift = 14
        counter += 1
    index = qwerty[letter]
    total = index + shift
    if total > 25:
        total = total - 26
    result += numqwerty[total]
    
print(result)

Results in this:

>>> ================================ RESTART ================================
>>> 
Insert goldstinger string: zm bakcag ova ohvor jezb x fzyt cbk kcl xsg wzanjgn kwxzb vsn ijgnxjb
if anyone can crack this i will pay you one hundred thous and credits

Congratulations reuben!

[Velocity]

Or how I more-or-less brute forced it:
http://pastebin.com/PXeC4pwe

(I looped a 0-25 shift on the entire thing before settling on 14)

[DossarLX ODI]

By the way the "thous and" part was NASTY. That extra space threw everyone in the code cracker chat off, haha.

[gold stinger]

Message has been sent to reuben_tate.

Here's the code I scribbled down in notepad that gave me the idea for this.

For the duration of this code, . = *

3this/code()is12almost5uncrackable.

thiscodeisalmostuncrackable

(12 x 3) / 5 = 7.2

7 - qwerty keyboard shift
5 degrees of separation
1 shift per degree

this1odei1almo1tunc1acka1le
3svgl/rhzp()gz12kmuuzs5fyeseeneymi.

if anyone can crack this I will pay you one hundred thousand credits.

11 + (15 x 4) + 8 - 13 - (5 / 3) = 64.33333continuous

-if+anyone-can11crack5this3I8will15pay+you4one()hundred/thous()and13credits.

ifanyonecancrackthisIwillpayyouonehundredthousandcredits
if_ny_ne_an_ra_kt_is_wi_lp_yy_uo_eh_nd_ed_ho_sa_dc_ed_ts
zm_ak_ag_va_hv_rj_zb_fz_tc_kk_lx_gw_an_gn_wx_bv_ni_gn_jb
-zm+_ak_ag-_va11_hv_r5j_zb3_8fz_t15c_k+k_l4x_g()w_an_gn/_wx_b()v_n13i_gn_jb.
-if+anyone-can11crack5this3I8will15pay+you4one()hundred/thous()and13credits.
-zm+bakcag-ova11ohvok5jezb3x8fzyt15cbk+kcl4xsg()wzanjgn/kwxzs()vsn13ijgnxjb.
-zm+bakcag-ova11ohvor5jezb3x8fzyt15cbk+kcl4xsg()wzanjgn/kwxzd()vsn13ijgnxjb.
-__-:__:__-:__--:__;_-_:__-:-__:_--_:_-_:_-_:_--_:__:__-:__:_--_:_--_:__:__-

12 - qwerty keyboard shift (IGNORED 64, ROT13)
3 degrees of separation
1 shift per degree

a=v
b=p
c=i
d=n
e=g
f=m
g=q
h=w
i=z
j=e
k=r
l=t
m=s
n=a
o=x
p=c
q=d
r=h
s=b
t=j
u=l
v=o
w=f
x=u
y=k
z=y

qwertyuiopasdfghjklzxcvbnm

[reuben_tate]

Originally Posted by gold stinger

I have sent 200,000 credits to the account of reuben_tate, for completing the code. Feel free to share your winnings as you see fit.

Currently sending something else for reuben in the form of PM. Will post answer to the code for everyone shortly.

Wow, that's more than what the message said <3

Anyways, I wanted to thank everyone in the FFR New Year's Codebreakers Skype group (yes, we took this threat very seriously): shoutouts to dossar, velocity, krazykitsune, danceflashrevo, and silvuh. Everyone pitched in a lot of ideas and shared what doesn't work, all of which made us progress closer to the answer.

Thanks to gold for putting this together, for the credits, and being good at being evil (see what I did there?)

[gold stinger]

I'll see you all next year with a harder code.





the amount of people kicking themselves in the pants right now for not noticing their keyboard sooner.

[gold stinger]

A code for you all to break, my humblest regards for those that still hunger for codebreaking.

أĕ└Ąḅᴘᴆ‡ᴒ↑→أḈ ́ ꜠←أĕᴘᴒĄ→ ́ ←↑ (גּ-ﭙ)

your hint is that it entirely uses a font called arial.

And you thought they couldn't get harder

[Velocity]

http://pastebin.com/uzFzbE9K

[popsicle_3000]

Originally Posted by reuben_tate

if anyone can crack this, I will pay you one hundred thousand credits

payment please

mfw I thought ben was asking people for help, with 100k reward

[gold stinger]

Velocity confirmed god



















were you secretly behind the solving of the enigma cipher