## Cipher Challenge

A forum for good logic/math puzzles.

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jianmin
Posts: 9
Joined: Sat Oct 01, 2011 9:14 pm UTC

### Cipher Challenge

I've got a cipher challenge for something at my school; the rules are that we're only allowed from sources outside of the school, so I'm here. The goal is to learn from the challenge, not just to solve it.

That all said, here is the challenge:

Code: Select all

fibonacci% 4D667171457A714581A69FEF39E2D51C9C649ECF278150617372A5DC

The hex in 28 ASCII characters (meaningless at the moment) :
MfqqEzqE.¦Ÿï9âÕ.œdžÏ'.Pasr¥Ü

Any thoughts?

Ankit1010
Posts: 135
Joined: Fri Feb 11, 2011 11:32 am UTC

### Re: Cipher Challenge

I'm assuming you know what the fibonacci sequence is... In this case, if you interpret the hex as decimal numbers correctly, you can see that it forms a kind of generalization of the fibonacci sequence.

Hint #1:
Spoiler:
Interpret every 2 hex numbers as one decimal number

Hint #2:
Spoiler:
The sequence is the fibonacci sequence with some extra terms. They are generated similarly to how the fibonacci terms are.

jianmin
Posts: 9
Joined: Sat Oct 01, 2011 9:14 pm UTC

### Re: Cipher Challenge

Thanks for the reply! I know what the fibonacci sequence is, though there may be variations of it that I'm unaware of.

I've been working through this for awhile over the past two days and have looked at the decimals several times; I guess I'm just not seeing the pattern you're describing.

The decimals:
Spoiler:
77,102,113,113,69,122,113,69,129,166,159,239,57,226,213,28,156,100,158,207,39,129,80,97,115,114,165,220
life means so much.

Ankit1010
Posts: 135
Joined: Fri Feb 11, 2011 11:32 am UTC

### Re: Cipher Challenge

Well... this is embarassing. So I wrote a program to convert the hex numbers to decimal, but ended up making a small bug that generated a kind of generalization of the fibonacci sequence, instead of converting the numbers correctly Hence, disregard everything I said in my previous post. I'll take a look at the sequence again when I have some time and see if I can figure it out.

ineon
Posts: 7
Joined: Tue Jan 31, 2012 11:23 am UTC

### Re: Cipher Challenge

Spoiler:
Subtracting the nth Fibonacci number from the nth number in that sequence gives the following numbers (for 0<=n<=11)
76 101 111 110 64 114 100 48 95 111 70 95
Convert these into ascii and you get
Leon@rd0_oF

It breaks down after n=11 as the next number is -176 which doesn't have a corresponding ascii representation.

sfwc
Posts: 221
Joined: Tue Mar 29, 2011 1:41 pm UTC

### Re: Cipher Challenge

ineon wrote:
Spoiler:
Subtracting the nth Fibonacci number from the nth number in that sequence gives the following numbers (for 0<=n<=11)
76 101 111 110 64 114 100 48 95 111 70 95
Convert these into ascii and you get
Leon@rd0_oF

It breaks down after n=11 as the next number is -176 which doesn't have a corresponding ascii representation.

Spoiler:
Since there are 256 possible ascii values, every number can be given an ascii value. First divide the number by 256, then take the remainder. To put it another way, add or subtract a suitable multiple of 256 to get the number into the range from 0 to 255. In this case, 256 - 176 = 80, which codes P. That's good, since the context suggests the next word could be Pisa.
Solution:
Spoiler:
If we carry on using the method ineon suggested, but adding or subtracting 256 whenever we have to to keep all the numbers in the range from 0 to 255, then we get the following message:

Leon@rd0_oF_PisA_LIbeR_AbAci

Ignoring the odd symbols being used to represent letters and the funny capitalization, we get the words `Leonardo of Pisa: Liber Abaci'

jianmin
Posts: 9
Joined: Sat Oct 01, 2011 9:14 pm UTC

### Re: Cipher Challenge

Wow, I had tried that, even have code that does the remainder for values > 256. But I started the fibonacci sequence where the first two values were 0 and 1 rather than 1 and 1. Changed it and it works beautifully. Thanks for the help!
life means so much.