question and answer thread.

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Originally posted by SeventhStar
hmmm i didn't say the "gif" image data did i?

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I did the same as Solarflare, and nothing worked.

But if we have to find which gif on the internet we need to try to convert, ...

By the way, at the end of the ".com" file, there is written: "the password is not here". The gif says "You'll find nothing here". That's not very helpful... :(

I don't have a clue where to start. I don't even know what we are looking for. And don't have time to look again and again at the post you've made, so I guees I'll stop here.

you obviously havent heard of the simplest image format in the world... the simplest of simple that doesnt even contain info about the image size, and colors...

find it and it will lead you to the solution :P

when i open the step02.com using binary.. i see lots of 'F' in the left and a few '@' in the right coloum..

don't tell me that makes up a beautiful picture.

Quote:

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Originally posted by wuyh
when i open the step02.com using binary.. i see lots of 'F' in the left and a few '@' in the right coloum..

don't tell me that makes up a beautiful picture.

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I've tried that as well but it doesn't look like anything. Maybe the editors I have are no the correct ones to see and ASCII picture, but it's just a big mess of nothing.

If we need a specific tool to be able to see it, it could be long to find.

i told you you need an exact image format... it's so simple that almost all image programs support it:

ACDSee
Photo Shop
Photo Paint
Paint Shop Pro
Irfan View

is that so hard? Think!
it's not hard once you have it

I don't know anything about image formats, and I don't know any format that contains no information about colour and size and still is an image, and that would be supported by an image editor, and I don't have any of these:
ACDSee
Photo Shop
Photo Paint
Paint Shop Pro
Irfan View
on my PC at work.

Does "Paint" support this image format (probably not as a bmp has colour information)?

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Originally posted by SeventhStar
you obviously havent heard of the simplest image format in the world... the simplest of simple that doesnt even contain info about the image size, and colors...

find it and it will lead you to the solution :P

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I obviously have not.

find this format then :p ask somebody (not me :D)

paint does not support it, so if dont have any image programs you can download irfan view (it's a GREAT program) from here:
http://www.irfanview.com/main_download_engl.htm

Actually, you don't even need an image editor if you have VC++ and know about the DIB format (the structures BITMAPINFO and BITMAPINFOHEADER come in handy) ;)

So rest assured that there is a clear message in step02. But I'm still stuck with step01...

haha :D
Actually i never knew that there was such a way for solving step 2. Actually the dib format isn't the one i meant but it can be solved with it... only why bother i only makes things more complicated.

There is a picture format that does NOT contain any info about the picture and viewers which open it let you choose the size and bits per pixel. This one must you search. The riddle is solvable with just a good picture editor :p

if you look at my posts you must find it

Quote:

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Originally posted by SeventhStar
Actually i never knew that there was such a way for solving step 2. Actually the dib format isn't the one i meant...

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No, of course not. But programatically creating a DIB is an easy way of displaying raw image data. :p

Sorry I am lost , What are we looking for? can anybody make me a brief resume?

What must we do with Step01.gif and Step02.com?


And what in **** sould I XOR with a key ?

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Originally posted by Doctor Luz
Sorry I am lost , What are we looking for? can anybody make me a brief resume?

What must we do with Step01.gif and Step02.com?


And what in **** sould I XOR with a key ?

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a brief resume is that you need to use both files to get the password. :D
dont know how :p and what

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Originally posted by gstercken
No, of course not. But programatically creating a DIB is an easy way of displaying raw image data. :p

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well you'll have a lot more problems with step1 if you stick to the dib format

Yes but sould I use any data from previous posts, number, key or something or this independent from the moment you posted the .zip?

Well, I've got the answer. Mythical, SeventhStar! ;) Do you want me to post it here, or should I rather PM it to you?

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Originally posted by Doctor Luz
Yes but sould I use any data from previous posts, number, key or something or this independent from the moment you posted the .zip?

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use whatever you want :D i gave a lot of hints in my las posts

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Originally posted by gstercken
Well, I've got the answer. Mythical, SeventhStar! Do you want me to post it here, or should I rather PM it to you?

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PM it to me and if tomorrow noone else has it you'll be the only winner :p (and genius;))

I don't find nothing in Step01.gif, I give up.

so gstercken was the only one to find the answer :) it was
"Mythos!!"
so, the solution:

"Raw image data", that's the solution: raw files hidden inside other files!

The first one, inside step01.gif, is an executable file: if you open the gif with an image editing tool, such as Paint Shop Pro or Gimp, and save it as a raw 256 color image, you just save all the single pixels you see on the screen as bytes inside the raw file. Finally, you will end with a .com which prints out the string

F7F064815376E13F

How should you know that you should consider it as a .com file? Well, there is a little hint inside the image: if you give a look at it (in its uncompressed format) you will se the words "run me".

Now keep in mind (or, better, pasted in a text file ;) these characters and work on the second file: if you run it file you will see the string "The pass is not here", but of course the pass IS there!
Just open the .com as a raw image: when you will be requested the size in pixel you wish to view it, choose width and height so that you keep the ratio of step01.gif. That is, if step01.gif is a 16000 pixel image of size 400x40, just choose for step02.com (which is 4000 bytes long) 200x20: you will end with a raw image containing this message:


"Hi! Now xor these values, [some pixels], with the key
you have, to find the correct password"


The values, reading from the pixels, are:

DEC: 186 137 016 233 060 005 192 030
HEX: BA 89 10 E9 3C 05 C0 1E

Now, let's xor the first sequence of hex values with the second one and let's see what happens:

F7 F0 64 81 53 76 E1 3F
BA 89 10 E9 3C 05 C0 1E
==============================
4D 79 74 68 6f 73 21 21


Which, in ASCII, is the string

M y t h o s ! !

here it is ;)
I hope gstercken wouldnt mind but i'll paste a part of his pm to me and ask you for your oppinion on this:

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After all, this is a software developer forum, and this kind of 'riddle' is exactly what I personally was always expected to see in the Q&A thread.

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do you all want me to proceed with such riddles or should we stick to the old facion way :D
let me here you :p

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Originally posted by SeventhStar
do you all want me to proceed with such riddles

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Yes! Yes!! Yes!!! Please go on - this is exactly the kind of challenge we all ;) :p like!

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Originally posted by SeventhStar
do you all want me to proceed with such riddles or should we stick to the old facion way :D
let me here you :p

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You can post that type of riddle if you like, but I cannot do them.

Quote:

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Originally posted by SeventhStar
How should you know that you should consider it as a .com file? Well, there is a little hint inside the image: if you give a look at it (in its uncompressed format) you will se the words "run me".

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I don't have these words: "run me". How do you get them? What is the uncompressed format of the file, how do you get it?

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Now keep in mind (or, better, pasted in a text file ;) these characters and work on the second file: if you run it file you will see the string "The pass is not here", but of course the pass IS there!

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I don't get that either. I have been able to find this string while looking at the binary content of the file, but it wouldn't run and give that string on my PC.

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do you all want me to proceed with such riddles or should we stick to the old facion way :D
let me here you :p

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Will they all require knowledge in image formatting? ;)

that are <= 1000^2000
:)
....

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Originally posted by hometown
that are <= 1000^2000
:)
....

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Okay... 1000^2000 = 10^6000 = 2 * 5x10^5999
every other number is odd, so the answer is 5x10^5999.

:)

I have an integer, could anyone tell me the number of operations I need to partition that number till it is a sum of all 1s
For example I have 5=4+1=2+2+1=2+1+1+1=1+1+1+1+1
I will get num(5)=4
:)
Thank you...

The way you're doing it seems to get just one new "+ 1" at each step. It means that you need N-1 steps.

num(N) = N - 1

I am sorry perhaps because my question mightnot be clear , you misundertood what I really want to ask...
When you partition the number, its parts must be a power of 2. Try again please...:)

Thanks...

I should point out at this stage that the question states 'any integer', and that can't be done, as negative integers can not be expressed as the sum of powers of positive numbers.

Also, I'm going to assume that when a number is initiately broken down, it is into the smallest number of powers of 2 possible, or else the answer is trivial (1 breakdown to turn everything into 1's).

For any power of two you break down, the total number of sub-breaks is equal to that number minus 1. Proof (sort of): 2 requires 1 break down, for each time you multiply this by two, you double the number of breakdowns and add 1, the initial breakdown. It looks kinda like a binary tree if you chart it.

For non-powers of two, the rules change a bit. You need the additional breakdown at the start which results in more than two powers of two created. The number of powers of two created is equal to the number of '1's in the binary representation of the number. Thus the total number of breakdowns is equal to the number of break downs for each power of 2, plus the initial break down. The number of additional breakdowns of all the powers created is equal to the sum of those powers (which is the original number) minus the number of powers created. Therefore:

num(N) = N + 1 - (number of '1's in binary representation of number)

so num(5) = 5 + 1 - 2 (there are two '1's in '101')

Unless there's a formula to determine the number of digits in a binary number based on it's decimal number other that just counting, that's pretty much it.

Sir, Could you tell me what made you think of using binary representation of that given number ? May I ask for that ? :)

Thank you....
Regards,

because of powers of 2 of course, forgive me for answering for him, but what else would someone do?

Yah, what Souldog said. Should I ask a question now?

Sure, go ahead.

Oh please not yet...:)
A.J.Gibson,
As what you have explained in your previous post, if my number is N there will be N' times that must be smaller than N to break down that N, is that right ?
Because what I have just been taught leads the result to something slightly different from that...Could you give me your answer ?...:)
Thanks...

P.S I am sorry just because I amnot really sure...

Quote:

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Originally posted by hometown
As what you have explained in your previous post, if my number is N there will be N' times that must be smaller than N to break down that N, is that right ?
Because what I have just been taught leads the result to something slightly different from that...Could you give me your answer ?...:)
Thanks...

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Are you saying my answer is wrong? If you looking for a solution that works with 0 or negative numbers, there isn't one, because you can't express a number less than 1 as the sum of positive numbers, and all powers of a positive number are positive.

Looking back at my answer, I see that I might not have been clear at one point: there are two formula's for the answer: num(N) = N-1 for powers of 2, and num(N) = N+1-(# of 1's in binary representation) for non-powers of 2. Is this what you meant? Or is one of my original assumptions wrong? I assumed the initial breakdown resulted in only powers of 2.

For example: say you're breaking down the number 23:
23
16+4+2+1 (this is the initial breakdown, the '+1')
8+8+4+2+1
8+4+4+4+2+1
4+4+4+4+4+2+1
4+4+4+4+2+2+2+1
4+4+4+2+2+2+2+2+1
4+4+2+2+2+2+2+2+2+1
4+2+2+2+2+2+2+2+2+2+1
2+2+2+2+2+2+2+2+2+2+2+1
2+2+2+2+2+2+2+2+2+2+1+1+1
2+2+2+2+2+2+2+2+2+1+1+1+1+1
2+2+2+2+2+2+2+2+1+1+1+1+1+1+1
2+2+2+2+2+2+2+1+1+1+1+1+1+1+1+1
2+2+2+2+2+2+1+1+1+1+1+1+1+1+1+1+1
2+2+2+2+2+1+1+1+1+1+1+1+1+1+1+1+1+1
2+2+2+2+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
2+2+2+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
2+2+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
2+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

That's 20 breakdowns, which is 23 + 1 - 4, the 4 being the sum of how many powers of 2 that 23 is (at minimum), namely 16, 4, 2, and 1.

Does that help?

A.J.Gibson
Please donot get angry...I think it is not important for the answer to be correct or incorrect because there are always unexpected countless things in between, it was me who made such an unclear question. The main purpose I had when asking you again and again was nothing but just try to understand what you were really thinking when you gave me such an answer and more than that, I would like to find out why I couldnot get the answer that was the same as yours, which might partially upset you... Now I understood...Donot get angry at me hah ?:)
Your answer is absolutely correct...
What I have learnt,however, is different from the detailed result you offered and explained here...Because everytime I partition a number, I have only one choice, which means I am allowed to break only one number into its binary components at one time. Therefore, if I want to do that with another number that is next to the one I have just "devided", I will have to go back to its original position.
For example, in your case:
23
16 4 2 1
16 4 1 1 1
16 2 2 2 1
16 2 2 1 1 1
16 2 1 1 1 1 1
16 1 1 1 1 1 1 1
8 8 4 2 1 // Go back to the second position *
8 8 2 2 2 1
8 8 2 2 1 1 1
8 8 2 1 1 1 1 1
8 8 1 1 1 1 1 1 1
8 4 4 4 2 1// Go back to the postion *
.....
This is repeated until
2 2 2 2 2 2 2 2 2 2 2 1
2 2 2 2 2 2 2 2 2 2 1 1 1
2 2 2 2 2 2 2 2 2 1 1 1 1 1
......
2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

So finally there are 74 steps to finish...
Finally, I have to say that I really like your explanation, it is natural in the way of reasoning, and... that really helps...

Thank you very much...:)

I'm not angry, hometown, I'm just confused. But now I've read your explanation, so I've moved on to bewildered :)

I'm going to ask a riddle now, because I think it's my turn. If it's not, ignore my riddle. My apologies if it's been done here already, I haven't read every single riddle here. Hope it's not to easy...I haven't been able to solve it, and I'll be kicking myself if it's something obvious.

Name a 10-digit number with 10 different digits such that taking any number of digits from the front of the number results in a number divisible by the number of digits. In other words, the first 2 digits form a 2 digit number divisible by 2. The first 7 digits form a 7 digit number divisible by 10. And so forth up to 10.

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Originally posted by A.J.Gibson
The first 7 digits form a 7 digit number divisible by 10.

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shouldn't this be by 7?

1296547830 is a (the number) that does it

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Originally posted by Elrond
1296547830 is a (the number) that does it

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why?

Sorry, the correct number is: 3816547290

The other will not work with 8...

Why? because it follows the rules.

3816547290->

10 * 381654729 = 3816547290
9 * 42406081 = 381654729
8 * 4770684 = 38165472
7 * 545221 = 3816547
6 * 63609 = 381654
5 * 7633 = 38165
4 * 954 = 3816
3 * 127 = 381
2 * 19 = 38
1 * 3 = 3


I just wrote a little program to find it ;)

Even without that, there are a few rules that would make it easier:

The last digit MUST be 0 (divisible by 10).
The 5th digit must therefore be 5 (first five divisible by 5, and 0 is gone).
You only care about finding the correct number up to 8 (if you follow the rule of having only different digits, you have 0 as the tenth, so when you end up with the remaining 9 digits, all numbers can be divided by 9, if you remember the rule that when the sum of the digits can be divided by 9, the number can be divided by 9).
Every 2nd digit must be an even digit (must be divisible by an even number).

I have not found any other rule yet, but it already reduces by a lot the number of possible answers... (about 2401 left if I am correct, but might not be as I am not that good with statistics).

Serves me right for trying to stump people with a question that can be solved by brute force. I was kinda hoping for something a little more elegant.

All right, Elrond, ask away.

Quote:

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Originally posted by A.J.Gibson
Serves me right for trying to stump people with a question that can be solved by brute force. I was kinda hoping for something a little more elegant.

All right, Elrond, ask away.

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Is there a more elegant solution? Probably yes, I'd like to know it. :)

I am about to leave and have no time to find a riddle now. Anyone who wants can ask one. Otherwise I'll try to find one tomorrow :cool:.

Fine, I'll ask one that can't be solved by brute force, hopefully none of you have read it before:

Arrange a group of 4 numbers such that with the addition of some symbols they form the two sides of an equal equation. For example: the numbers 3,4,6,9 can form 4*9=36. Okay? The numbers in question are: 2,4,7, and 9.

Too easy?

Do you want to try a riddle in the style of SeventhStar?

Well, you have to look for a password hidden in the attachment.

There is a 'poem' which will give you a hint for knowing what to do with the attachment.

I think this is very easy.

Read carefully and sorry for the bad english.

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Come here and read.
Here is a riddle again.
After finding the password,
no problems will remain.
Gurus and not gurus can try and
easier riddle will not find.
The first thing you should do is the
hint find.
Extend your mind,
search for the hint to victory.
Extend your mind,
code is the solution.
Observe as well as you can and
none of the hints will hide.
Delicate the hint is, so
you should read carefully.
Extend your mind,
look at this text.
Look at the attachment.
Obviously you are a
Winner...

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:D

Quote:

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Originally posted by A.J.Gibson
Fine, I'll ask one that can't be solved by brute force, hopefully none of you have read it before:

Arrange a group of 4 numbers such that with the addition of some symbols they form the two sides of an equal equation. For example: the numbers 3,4,6,9 can form 4*9=36. Okay? The numbers in question are: 2,4,7, and 9.

Too easy?

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9 = 7 + 4 - 2

Quote:

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Originally posted by souldog
9 = 7 + 4 - 2

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Hmmm...I must have forgotten something, because that's not the original answer.

Well, you did it using 2 symbols (plus and minus), can anyone do it with less? I can :) In fact, I can do it without using any symbols at all...

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Originally posted by A. J. Gibson
Well, you did it using 2 symbols (plus and minus), can anyone do it with less? I can In fact, I can do it without using any symbols at all...

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Well, I'm not sure about what counts as a symbol here, but
47 = 2 (mod 9)
if that's the kind of thing you were thinking...

'mod' counts as symbol. In fact, forget symbols completely - just have an equal sign.

You know, this riddle would have been much better if I had gotten it right in the first place.