Gimre's riddles #01

General Discussion
#1

.rules.

Riddles closed 27/05 14:00 GMT

There are 6 riddles/tasks of different difficulty.

Every task has it’s own reward AND address where to send an answer.
Please DO NOT post answers here.

First CORRECT answer wins, so answers don’t have to be encrypted.
Rewards will be sent back to address from which answer came.

There might bonus riddles (but I don’t promise anything). Keep observing OP.

Have fun!

1. Book title (1000 XEM)

CLAIMED by ndxrv4-767uox-c2cy3y-irlfmb-twixxs-v7gqyo-welc

answer: 450b0896848a992f037eb4d34a0f34734d40f8001fdebeba31989afaf80df951
reward: bd4a5a2ba1773b620ee709f168e88bf3e93fa3cf9ae27abdea8e03d2ed7439cb

A friend has send me quote from the book. Can you tell me from which one?

Iurp wkh gudzhu ehvlgh wkh vlqn Mrh Fkls jrw d vwdlqohvv vwhho nqlih;
zlwk lw kh ehjdq vbvwhpdwlfdoob wr xqvfuhz wkh erow dvvhpeob ri klv
dsw’v prqhb-jxoslqj grru.

”L’oo vxh brx,” wkh grru vdlg dv wkh iluvw vfuhz ihoo rxw. Mrh Fkls vdlg,
”L’yh qhyhu ehhq vxhg eb d grru. Exw L jxhvv L fdq olyh wkurxjk lw.”

2. Thermoptic camouflage (8500 XEM)

CLAIMED in second attempt of ndxrv4-767uox-c2cy3y-irlfmb-twixxs-v7gqyo-welc

answer: 703e522766b716283364476b1ee55527a3a5b2276779b218a2cd012990e64c16
reward: 0d0a8ac173d3f963211ba3758f5de0ab01fd57a2217134e822c1860a203ab6e5

It just doesn’t sound right.

https://drive.google.com/file/d/0B_te3HSCdxPVWTVEOFI1eUZoZFk/view?usp=sharing

3. Who’s the hero (1500 XEM)

CLAIMED in second attempt of nae66a-mg4xi3-wqtjt7-hx4a7a-grhp5x-6waaap-z4bx

answer: 1981e29ea0efbd8627c6e0a8ca97b1d2e09e3db4a63d8d681214c23923d6b0f6
reward: 91876bdf1c93f5500fb650cd703b66f3219cdec707db0cc51b941ee9bdc5a2bc

I’m not sure who is it about. Perhaps you know?

d0 92 20 d0 b1 d0 b5 d0 bb d0 be d0 bc 20 d0 bf
d0 bb d0 b0 d1 89 d0 b5 20 d1 81 20 d0 ba d1 80
d0 be d0 b2 d0 b0 d0 b2 d1 8b d0 bc 20 d0 bf d0
be d0 b4 d0 b1 d0 be d0 b5 d0 bc 2c 20 d1 88 d0
b0 d1 80 d0 ba d0 b0 d1 8e d1 89 d0 b5 d0 b9 20
d0 ba d0 b0 d0 b2 d0 b0 d0 bb d0 b5 d1 80 d0 b8
d0 b9 d1 81 d0 ba d0 be d0 b9 20 d0 bf d0 be d1
85 d0 be d0 b4 d0 ba d0 be d0 b9 2c 20 d1 80 d0
b0 d0 bd d0 bd d0 b8 d0 bc 20 d1 83 d1 82 d1 80
d0 be d0 bc 20 d1 87 d0 b5 d1 82 d1 8b d1 80 d0
bd d0 b0 d0 b4 d1 86 d0 b0 d1 82 d0 be d0 b3 d0
be 20 d1 87 d0 b8 d1 81 d0 bb d0 b0 20 d0 b2 d0
b5 d1 81 d0 b5 d0 bd d0 bd d0 b5 d0 b3 d0 be 20
d0 bc d0 b5 d1 81 d1 8f d1 86 d0 b0 20 d0 bd d0
b8 d1 81 d0 b0 d0 bd d0 b0 20 d0 b2 20 d0 ba d1
80 d1 8b d1 82 d1 83 d1 8e 20 d0 ba d0 be d0 bb
d0 be d0 bd d0 bd d0 b0 d0 b4 d1 83 20 d0 bc d0
b5 d0 b6 d0 b4 d1 83 20 d0 b4 d0 b2 d1 83 d0 bc
d1 8f 20 d0 ba d1 80 d1 8b d0 bb d1 8c d1 8f d0
bc d0 b8 20 d0 b4 d0 b2 d0 be d1 80 d1 86 d0 b0
20 d0 b8 d1 80 d0 be d0 b4 d0 b0 20 d0 b2 d0 b5
d0 bb d0 b8 d0 ba d0 be d0 b3 d0 be 20 d0 b2 d1
8b d1 88 d0 b5 d0 bb 20 d0 bf d1 80 d0 be d0 ba
d1 83 d1 80 d0 b0 d1 82 d0 be d1 80 20 d0 98 d1
83 d0 b4 d0 b5 d0 b8

4. Bit too small (25000 XEM)

SOLVED, but giving a chance to other participants!

CLAIMED by nae66a-mg4xi3-wqtjt7-hx4a7a-grhp5x-6waaap-z4bx

answer: (hopefully will be below)
reward: 29f47948018e11b9d7961ce34e894ae7b3dd91ccfdc13d41c6dce329fb83a915

I was given following public key, can you tell me what is the private exponent?

-----BEGIN PUBLIC KEY-----
MB8wDQYJKoZIhvcNAQEBBQADDgAwCwIGAS4Ye9atAgEH
-----END PUBLIC KEY-----

5. Who is SHE? (12000 XEM)

CLAIMED by nae66a-mg4xi3-wqtjt7-hx4a7a-grhp5x-6waaap-z4bx

answer: beb91cc02efa3314648116959a8285543866ff987ca466ff85d70f22f9648f01
reward: 3b41b6091f602ecf33938703df32ad2f47c35706e8a886e19d488780975769f3

We know who they are, but it has come to our attention that she might
have valuable knowledge…

4 Likes
Gimre's riddles #02
#2

http://chain.nem.ninja/#/transfer/e9153d203b8e6faf0d848a01a8f47c111c59e75fe0c7ad5b99608a7cafcd5e39

That would be a little too obvious, wouldn’t it? ^^

#3

dang you guys are fast!

how’d they get wang xiaoyun?

#4

I have no idea, descriptions will be posted, after whole thing will be closed…

1 Like
#5

Where is the 6th riddle? ^^

2 Likes
#6

So although I was too slow for riddle number two , I will clarify it :

The message decodes to : В белом плаще с кровавым подбоем
which tranlsates to : In a white cloak with blood-red lining

This is a sentance from the famous novel by Mikhail Bulgakov:

The two main characters in the book are Jesus and Pontius Pilatus, but apparently the hero in the book was the latter :slight_smile:

#7

6th (and possibly 7th) were supposed to be during the weekend, but I’m not sure if I manage to do that.

#8

@Xpedite task stated: “I’m not sure who is it about. Perhaps you know?” that question refers to the fragment, fragment was about Pontius Pilate :slight_smile:

1 Like
#9

:slight_smile: Well, I’ve heard from the book before but never read it.
Now I know what to read during my summer holiday :wink:

#10

So… I took the picture and inverted colors:

Okay, I’m looking for an asian woman.

Now I took the original and put it into Google Image search to find out the context:
LOOONG URL

Now I knew that those guys on the picture are some famous cryptographers.

So I thought ‘Let’s look for an asian female cryptographer’ and ended up on this Wikipedia page where I looked up all asian names. Then I looked up ‘Xiaoyun Wang’ on Google: LINK

Oh, you have seen that woman before :smiley: There she is…

2 Likes
#11

So this one was not so easy without knowledge about RSA encryption :wink:

First I looked up some phrase like ‘get private key from public key’ on google and read about how to crack insecure RSA keys.

The final formula for what we are looking for is:

d = e^-1 mod phi(N)

Then I booted up my Linux VM and made my first steps to the solution:

# openssl rsa -pubin -in pub.txt -modulus -text
Public-Key: (41 bit)
Modulus: 1297490892461 (0x12e187bd6ad)
Exponent: 7 (0x7)
Modulus=12E187BD6AD
writing RSA key

So the Modulus (N) is 1297490892461 and the public exponent (e) is 7.

We now have e and need to calculate phi(N):

phi(N)=(p-1)*(q-1)

To get p and q you have to factorize the Modulus (fortunately there is a website doing this for you :wink: ):
http://www.factordb.com/index.php?query=1297490892461

Which tells you p=1000003 and q=1297487.

Putting all this together (I used a tool called CrypTool) you will have

d = 370711027135

… which is the solution :wink:

2 Likes
#12

I copied the text into a text editor and searched for small words which i then tried to guess. I thought that “wkh” could be “the” and wrote down which letter translated to which right letter. Then i replaced the letters in the text editor with the right ones that i found out which lead to more and more words that i could guess. I repeated these steps until i had one complete quote.

I searched for the quote on google and found the book in which it appeared - Ubik. :slight_smile:

IMG_2269.JPG3024×4032 1.55 MB

Here the notes i took while solving the riddle :smiley:

1 Like
#13

Okay - this one was was a little harder. I first searched for the name of the song with Shazam on my phone. The song name was 19-2000 by Gorillaz. I wasn’t quite sure what was required so i thought i’d take a shot and sent the song name as my solution, but that would have been to easy indeed, so i kept on searching.

The next thing i did was comparing the track provided by @gimre to the original after which i found some little differences. I thought that maybe there were two separate tracks laid on top of each other. I downloaded Audacity to look if i could somehow detect and take apart the different song layers (i had no clue if that was possible or not). After playing with Audacity for a while i came across a function which showed the spectrogram of the sound track and BOOM - i saw a hidden message :smiley:

Bildschirmfoto 2016-05-26 um 18.33.14.png923×303 143 KB

It read “Secret NEM” which had to be the solution of the riddle - i got a little lucky with this one :smiley:

These riddles were really cool - thank you @gimre for providing the interesting game! :smiley:

2 Likes
#14

All these solutions at such a short time tells me NEM community is pretty intelligent and creative :wink:

Now that we have a benchmark , can we (temporary) delete this thread and see how other crypto community’s perform :wink:

#15

Doesn’t get any cooler than that IMO !

1 Like
#16

You can change the settings in audacity prefs. By default it uses pretty small window (so that calculation won’t take long), here’s larger Hann window:

Pasted image881×657 826 KB

Also SOX can produce very nice spectrograms

spectrogram.png943×593 310 KB

2 Likes
#17

P.S. There’s a very simple ‘key’ behind that substitution ^^

1 Like