[Guest]

A master criminal has broken into a jeweller`s strong room which has the walls lined with several thousand storage boxes protected by individual four digit combination lock codes.. Each storage box contains small bags of precious stones (diamonds etc.) so the total value of all the contents of the locked boxes in this room will be several million of ££s .

Each box is labelled with a letter and four digits and the thief has inside information and knows that the combination lock code for each individual box can be mathematically calculated using the Box Number .He also knows that the only digits used for the code will be 1 , 2 , 3 , 4 and 5.

Eg ...........Box No J 0023 has a Combination Lock Code of 1215

However , each box has its Combination Lock Code engraved on the INSIDE and fortunately several of these boxes are empty and have been left OPEN . This means that the criminal can now list each open Box Number with its own Security Code , and , using this information will attempt to work out the method of calculating any Security Code from its Box Number.

His inside information has told him that there is one particular box containing the master list of what is contained in each box in the strong room...Box J 6383 ... so this is the box to open first.

Here is the list of open BOX numbers with their security CODES engraved on the inside.

Can you work out how the codes can be calculated and explain the method used to determine the combination lock code of Box J 6383 ?

.BOX = CODE..... BOX = CODE..... BOX = CODE

J 0023 = 1215.... E 2182 = 5225.... Z 6163 = 1445

H 0124 = 4441.... Z 2271 = 1552 .... J 6242 = 3545

C 0138 = 4513.... K 2380 = 5322 .... A 6322 = 4525

Y 0209 = 5554.... A 2408 = 4124 .... K 6481 = 3513

E 0215 = 3235.... Z 2494 = 4221 .... J 6519 = 1525

H 0262 = 1234.... Z 2812 = 5332 .... K 6620 = 3341

R 0448 = 5132.... A 2994 = 4425 .... J 6709 = 5355

R 0464 = 2135.... C 3010 = 5414 .... E 6999 = 4515

W 0496 = 2125 .... Z 3229 = 3321 .... A 7238 = 3222

R 0545 = 2545 .... D 3631 = 1152 .... E 7402 = 5454

A 0704 = 5122 .... L 3865 = 2545 .... A 7535 = 3555

A 0859 = 4453.... A 4089 = 4455 .... A 7713 = 5542

R 0871 = 5545 .... Z 4108 = 2414 .... A 7962 = 4431

Y 0926 = 4254 .... E 4111 = 3222 .... U 8056 = 5543

Z 1043 = 4455 .... A 4166 = 3325 .... E 8108 = 2332

V 1044 = 3442 .... K 4431 = 3552 .... E 8121 = 2522

U 1095 = 2245 .... A 4476 = 4343 .... E 8250 = 4531

U 1096 = 1142 .... S 4705 = 4133 .... Z 8596 = 1335

V 1192 = 4521 .... A 4707 = 2455 .... E 8391 = 3232

R 1244 = 1355 .... Z 4725 = 2325 .... A 8844 = 5432

H 1381 = 4255 .... J 5264 = 4524 .... A 8930 = 4545

E 1390 = 4251 .... Z 5271 = 1225 .... A 9024 = 5153

R 1479 = 1523 .... J 5362 = 5535 .... K 9139 = 3443

K 1533 = 5244 .... E 5567 = 3123 .... K 9356 = 2424

J 1853 = 5322 .... Z 5970 = 5452 .... A 9751 = 3242

A 6053 = 3245 .... Z 9800 = 3134

Good luck !

[Guest]

This puzzle is similar to the one I posted in August 2008.It was set by a Maths Lecturer as a holiday brain teaser for a group of University engineering students in order to keep their brains active over last Christmas (2008 ) holiday period !! (In theory at least)

I have spent time on this and can only solve part of it ....so I invite you to help out.

I am not certain but I believe that the LETTERS used are not involved in the final answer. This was a hint that I was given in order to save me wasting too much time and becoming disheartened with trying to solve the full puzzle. The trouble is that I do not know whether this hint is a "double bluff" and the letters are used... !!All I can say is that my progress so far has not used any of the letters

[Guest]

It's certainly difficult to calculate while The Ramones are pounding into your head....

Still working on it....

[Guest]

A master criminal has broken into a jeweller`s strong room which has the walls lined with several thousand storage boxes protected by individual four digit combination lock codes.. Each storage box contains small bags of precious stones (diamonds etc.) so the total value of all the contents of the locked boxes in this room will be several million of ££s .

Each box is labelled with a letter and four digits and the thief has inside information and knows that the combination lock code for each individual box can be mathematically calculated using the Box Number .He also knows that the only digits used for the code will be 1 , 2 , 3 , 4 and 5.

Eg ...........Box No J 0023 has a Combination Lock Code of 1215

However , each box has its Combination Lock Code engraved on the INSIDE and fortunately several of these boxes are empty and have been left OPEN . This means that the criminal can now list each open Box Number with its own Security Code , and , using this information will attempt to work out the method of calculating any Security Code from its Box Number.

His inside information has told him that there is one particular box containing the master list of what is contained in each box in the strong room...Box J 6383 ... so this is the box to open first.

Here is the list of open BOX numbers with their security CODES engraved on the inside.

Can you work out how the codes can be calculated and explain the method used to determine the combination lock code of Box J 6383 ?

.BOX = CODE..... BOX = CODE..... BOX = CODE

J 0023 = 1215.... E 2182 = 5225.... Z 6163 = 1445

H 0124 = 4441.... Z 2271 = 1552 .... J 6242 = 3545

C 0138 = 4513.... K 2380 = 5322 .... A 6322 = 4525

Y 0209 = 5554.... A 2408 = 4124 .... K 6481 = 3513

E 0215 = 3235.... Z 2494 = 4221 .... J 6519 = 1525

H 0262 = 1234.... Z 2812 = 5332 .... K 6620 = 3341

R 0448 = 5132.... A 2994 = 4425 .... J 6709 = 5355

R 0464 = 2135.... C 3010 = 5414 .... E 6999 = 4515

W 0496 = 2125 .... Z 3229 = 3321 .... A 7238 = 3222

R 0545 = 2545 .... D 3631 = 1152 .... E 7402 = 5454

A 0704 = 5122 .... L 3865 = 2545 .... A 7535 = 3555

A 0859 = 4453.... A 4089 = 4455 .... A 7713 = 5542

R 0871 = 5545 .... Z 4108 = 2414 .... A 7962 = 4431

Y 0926 = 4254 .... E 4111 = 3222 .... U 8056 = 5543

Z 1043 = 4455 .... A 4166 = 3325 .... E 8108 = 2332

V 1044 = 3442 .... K 4431 = 3552 .... E 8121 = 2522

U 1095 = 2245 .... A 4476 = 4343 .... E 8250 = 4531

U 1096 = 1142 .... S 4705 = 4133 .... Z 8596 = 1335

V 1192 = 4521 .... A 4707 = 2455 .... E 8391 = 3232

R 1244 = 1355 .... Z 4725 = 2325 .... A 8844 = 5432

H 1381 = 4255 .... J 5264 = 4524 .... A 8930 = 4545

E 1390 = 4251 .... Z 5271 = 1225 .... A 9024 = 5153

R 1479 = 1523 .... J 5362 = 5535 .... K 9139 = 3443

K 1533 = 5244 .... E 5567 = 3123 .... K 9356 = 2424

J 1853 = 5322 .... Z 5970 = 5452 .... A 9751 = 3242

A 6053 = 3245 .... Z 9800 = 3134

Good luck !

Is there no-one taking up the challenge?

[Guest]

I have spent time on this and can only solve part of it ....so I invite you to help out.

Can you share what you've found?

[Guest]

Can you share what you've found?

I have only managed to work out how to calculate the first code digit.

See spoiler for clues

From the list of boxes/codes , if you compile a list of boxes all with the same FIRST CODE DIGIT , what can you deduce?

eg...using a first code digit of 4

H 0 124 = 4 513

A 0 859 = 4 453

Z 1 043 = 4 442

Y 0 926 = 4 254

A 8 930 = 4 545

What combination of digits in the Box number look the same. (or at least similar ) ..to give the first code digit of 4?

Compare this with other first code digits ( of 1 2 3 4 5 )

The hint above is in the spacing of the numbers

Bear in mind that the code can only be a maximum digit of 5 .....so what do you do when any combination or calculation of numbers gives a number HIGHER than 5 ?

If you want the answer to this hint then I`ll post it later today under a Spoiler

Edited December 1, 2009 by redrooster

[Guest]

This puzzle is similar to the one I posted in August 2008.

If anyone is interested here is the link to this first Codebreaker puzzle from last year....complete with solution.

Click on the above link

If you wish to read through this first puzzle it will give an idea of the explanation of how that particular code was cracked.

Parts of it were rather difficult (for me at least! ) but the solution was eventually worked out by fellow BrainDen members.

The "multiply by 7 " idea ( where the resulting " units" digits were used as the code answer for part of the calculation ( ie ..the series 7 4 1 8 5 2 9 6 3 0 ) was interesting

(Also considered were the "multiply by 3 " idea in a similar manner ( ie :the series ...3 6 9 2 5 8 1 4 7 0)

Both of these "multiply by ..." ideas are useful because the first ten multiples(of both 3 and 7 ) cover the full range of numbers from 1 to 0 ...so they are an easy substitute when creating any simple code

So you can see some of the methods possible that can be used to break a code of this type.

However I would suspect that this current puzzle will , more than likely, be completely different !

Edited December 1, 2009 by redrooster

[Guest]

Is there anyone out there taking up this challenge?

[Guest]

Hint /Explanation / Part answer for the first Code Digit

Using these samples from the list

H 0 124 = 4 513

A 0 859 = 4 453

Z 1 043 = 4 442

Y 0 926 = 4 254

A 8 930 = 4 545

If you add the second+third+fourth digits of the Box Number ...what is common ?

From the above

1 + 2 + 4 = 7

8 + 5 + 9 = 22

0 + 4 + 3 = 7 ( or 17 if you count 0 as 10 )

9 + 2 + 6 = 17

9 + 3 + 0 = 12 ( or 22 )

Bearing in mind that the Code digits go no higher than 5 ,

then all the above totals are 2 more than 5 (or multiple of 5 )

(ie 7 or 12 or 17 or 22 )

So the common point on the left (the Box Numbers ) is 2

The common point on the right (Code number ) is the first code digit of 4

Therefore if adding the second + third + fourth digits of the Box number produces a 2 ( or a 7 or 12 or 17 or 22 etc. ) then the first Code Digit will be a 4

Try it for other Box Numbers + Code Numbers to confirm that this statement is true......then deduce the rest of the total values

ie...What is the first Code Digit if they total 1 or 3 or 4 or 5 ?

Edited December 9, 2009 by redrooster

[Guest]

$50 (USD ) PRIZE OFFERED

I have been trying to solve this problem for the past couple of months and naturally I am really keen to find a working solution. I have also made enquiries but cannot find any hints or answers from the original source

So I in order to provide an incentive for an answer , I am willing to offer a prize for the first person to fully solve this problem providing there is a full explanation of how the calculations are made and this will work on all the Box Number / Code Numbers listed above and I can fully understand how it is calculated .....so no guessing !

I have already solved Code Digit ONE so will pay $10 (USD or equivalent) for each of the solutions for Code Digits TWO and THREE and $20 (USD or equivalent ) for Code Digit FOUR.

I will also add an extra $10 to Code Digit FOUR if the final solution is found before the New Year (Jan 1 2010)

The decisions for awarding each prize will be mine...but I will be fair + honest and could share it between successful members working on each answer.

Please let me know if you are interested in helping me with this challenge

All payments will be made using PayPal.

[Guest]

There aren`t many people working on this puzzle .

I think I`ve solved the first code digit part (even though you have virtually given the answer in your last spoiler.)

[Guest]

is that correct?

if it is then I'll carry on with the next code digit.

add the 2nd + 3rd + 4th box digits

subtract any multiple of 5

this leaves total digit of 1, 2, 3, 4, or 5

if this total digit is 1 then the first code digit will be 5

if 2 = 4

if 3 = 3

if 4 = 2

if 5 = 1

this has been proven to work ok by checking all the box numbers in the list

an example

A6053 = 3245

0+5+3=8 (0 can be value 0 or 10 it doesn`t make any difference)

subtract 5 =3

From list 3=3 which is the correct 1st code digit

Edited December 12, 2009 by wirewheels

[superprismatic]

is that correct?

if it is then I'll carry on with the next code digit.

add the 2nd + 3rd + 4th box digits

subtract any multiple of 5

this leaves total digit of 1, 2, 3, 4, or 5

if this total digit is 1 then the first code digit will be 5

if 2 = 4

if 3 = 3

if 4 = 2

if 5 = 1

this has been proven to work ok by checking all the box numbers in the list

an example

A6053 = 3245

0+5+3=8 (0 can be value 0 or 10 it doesn`t make any difference)

subtract 5 =3

From list 3=3 which is the correct 1st code digit

Your first few lines should be:

add the 2nd + 3rd + 4th box digits

subtract any multiple of 5

this leaves total digit of 0, 1, 2, 3, or 4

if this total digit is 1 then the first code digit will be 5

if 2 = 4

if 3 = 3

if 4 = 2

if 0 = 1

[Guest]

judging by what the last few comments have been, i'm guessing that the code begins witha 2...

and i'm 13. surely you guys should have worked that out ages ago considering you're so clever and all. anyway...

bye guys... xx

[Guest]

sorry superprismatic... i didn't realise you had already said that... xx

[Guest]

OBSERVATIONS..This is assuming that the prefix LETTER is not part of any calculation

(i) Changing the FIRST Box Digit affects the SECOND , THIRD & FOURTH Code Digits only (and not the FIRST Code Digit)

So this means that the FIRST Code Digit is dependent on the SECOND,THIRD & FOURTH Box Digits only

Z 5271 = 1225

Z 2271 = 1552

(ii) Changing the SECOND Box digit affects ALL four Code Digits

V 1044 = 3442

R 1244 = 1335

(iii ) Changing the THIRD Box digit affects ALL four Code Digits

S 4705 = 4133

S 4725 = 2325

(iv) Changing the FOURTH Box digit affects ALL four Code Digits

S 4705 = 4133

A 4707 = 2455

note...the examples below show that it only affects three of the four Code Digits

Z 1043 = 4455

V 1044 = 3442

U 1095 = 2245

U 1096 = 1142

From the above (ii), (iii) and (iv) it looks as though Code Digits TWO , THREE and FOUR will involve using all the four Box Number digits in any calculations

(v) Different Boxes having the SAME CODE

E 4111 = 3222

A 7238 = 3222

A 4089 = 4455

Z 1043 = 4455

[superprismatic]

Here's my work so far:

I decided to write a program to see if any code digit

depends on a particular subset of the box identifier

positions (the letter with the four digits). So, I

looped over all subsets of Box identifiers. For example,

while processing the subset consisting of the first, third,

and fourth box identifier positions, I would only

consider the E39 from the Box identifier E1390. Then, I

would look thru all other Box identifiers which had

an E, 3, and 9 in any order. Then I would count which

positions in the code for the 2 boxes did not change.

If I found that a particular code position changed when

the subset of the first, third, and fourth positions

in two Box identifiers were the same, I would conclude

that that particular code position did not depend solely

depend on the first, third, and fourth positions.

I did this for all nonempty subsets of Box identifier

positions, all pairs of the 77 box identifier & code

instances, and all code positions.

I found that there were 11 instances where positions

3, 4, and 5 of one Box identifier had the same subset

of numbers as that of another Box identifier. In all

11 cases, their first code numbers matched. We already

knew that the first code position depended on the

numbers in positions 3, 4, and 5, however, it was a

good test of my program that it found that. The bad

news is that I found no other code position for which

this kind of agreement happened once, let alone 11 times,

with any subset of Box identifier positions.

It seems to me, then, that the individual code positions

(except the first) are not functions of any subset

of Box identifier positions. So, I think we should start

looking at ways to produce the last three digits

as a single entity from the Box identifiers.

I hope someone duplicates my work, as this would validate

(or possibly invalidate) my claim. I will gladly offer

any help or answer any questions about this problem.

Redrooster: be assured that some of us are working on this

problem, but it's difficult to report on negative results.

[superprismatic]

Redrooster, please check your original post for typos in the data. If you find any, I'd like to re-run my program. After all, the way I did things, one bad data point would spell disaster!

[Guest]

Redrooster, please check your original post for typos in the data. If you find any, I'd like to re-run my program. After all, the way I did things, one bad data point would spell disaster!

Yes..the data has been checked and is exactly as I received it

This is the fifth codebreaker puzzle that I`ve attempted from this source and have to say that all the previous ones have utilised working out single individual digits in the Code Number....so it`s an interesting theory that you are pursuing so I will be intrigued to see if it produces any logical answers.

On this particular puzzle , I worked out the first code digit fairly quickly but must admit that I am rather baffled with the other three.(and have spent a long time thinking about it !)

[Guest]

Is there anyone working on this puzzle?

If so , would you like to share any useful theories ....as I seem to have reached a brick wall...and cannot see (or theorise ) a way around it.

[Guest]

yea, im working on it few days but figured almost nothing just things i think everyone knows, f.e. it can be possible to use code digits you already know to find out another code digit or something, then i just dont believe that letters are for nothing there, i thought too about using another "reverse" thing (i mean 1=5 2=4 3=3 4=2 5=1 thing,) maybe some box numbers reversing (like A 0859 "reversed" is A 1362 when you use 1=0 2=9 3=8 4=7 5=6 etc.)and then somehow add/subtract it but no success right now. hope we will figure something out together coz i hate unsolved riddles (sry for my english )

[Guest]

k i have another theory. cant be this code/riddle just impossible to solve ? without answer ? i think if its true that teacher wanted to give to his students something to keep their brains active during holidays, he didnt want something that one can solve during few minutes, maybe hours, then if someone cant figure it out in short time, he will call to someone who knows the answer to find out and then all "keep-brains-active" thing is pretty pointless. Instead he would make something that cant be solved. This way he can keep all students trying to solve something which cant be solved whole christmas(coz they dont want to fail or something). Keep their brains active. And i think that's what he wanted. If someone asks what about that first solved code digit, well, it can be some kind of bluff, i would believe more in answer(and i did) if i could get part of it, than if i couldnt figure anything (imagine if u get f.e. 25xx box code answer and u need to get just 2 more, u wouldnt think about possibilty that last two numbers are just random and there is no way to find them, but when you try to solve it few days and still get xxxx

, well i would rly let it be). Well, i will still be doing this code till i wont get some 100% answers (dont care if its Box J 6383 code number or answer "It cant be solved.") but try to think about this "solution" too. I think its possible just like its possible that we cant just figure right combinations. (plz if you know someone who can say that it has answer consisted of 4 box code digits, say it, i will be so glad to hear it )

[Guest]

k i have another theory. cant be this code/riddle just impossible to solve ? without answer ? i think if its true that teacher wanted to give to his students something to keep their brains active during holidays, he didnt want something that one can solve during few minutes, maybe hours, then if someone cant figure it out in short time, he will call to someone who knows the answer to find out and then all "keep-brains-active" thing is pretty pointless. Instead he would make something that cant be solved. This way he can keep all students trying to solve something which cant be solved whole christmas(coz they dont want to fail or something). Keep their brains active. And i think that's what he wanted. If someone asks what about that first solved code digit, well, it can be some kind of bluff, i would believe more in answer(and i did) if i could get part of it, than if i couldnt figure anything (imagine if u get f.e. 25xx box code answer and u need to get just 2 more, u wouldnt think about possibilty that last two numbers are just random and there is no way to find them, but when you try to solve it few days and still get xxxx

, well i would rly let it be). Well, i will still be doing this code till i wont get some 100% answers (dont care if its Box J 6383 code number or answer "It cant be solved.") but try to think about this "solution" too. I think its possible just like its possible that we cant just figure right combinations. (plz if you know someone who can say that it has answer consisted of 4 box code digits, say it, i will be so glad to hear it )

I am fairly certain that this will be solvable. This was set as a problem for University Engineering students who were assured that there was a working answer

If you re-read the messages at the beginning , you will find a link to earlier Codebreaker puzzles from the same source. At one stage, like you , I thought that these earlier puzzles were "unsolvable" ...but some of the more industrious members on this forum found working solutions . Take a look and see if you can understand the answers found.A couple of the methods used are quite ingenious

So for this one , then as we are finding , it won`t be easy, but there will be a working answer.

I prefer puzzles like this where you can spend a lot of time trying to work them out.It can certainly be more frustrating than Sudoku , but once a solution is found , then the achievement felt is a lot higher .

( During World War 2 , at Bletchley Park , they had no choice but to keep trying to crack the codes so I like to think the same about this one.!)

Don`t forget that I`m offering prizes for working solutions.

[Guest]

Here are the links to the previous , similar Codebreaker Wanted problems from the same source.....and they looked unsolvable as well ....until they were solved by members of this forum.

Read through them and see what you think

Edited December 31, 2009 by redrooster

[Guest]

I have a theory:

Let each position be named thus: BL B1 B2 B3 B4 = C1 C2 C3 C4

Now, we know that there is an interaction in B2, B3 and B4 that produces C1 by using a sum to determine the relative anti-position of the first element of the code.

My theory is that instead of an interaction in B1, B2, B3 and B4 producing C2, that it is instead an interaction between either B1 and C1, or B1, B2, B3, B4 and C1, with the most likely being an interaction between B1 and C1. I've been evaluating squares, square roots, multiples, divisors, sums, differences, everything I could think of, and I haven't found anything yet. Still working on it, though, until I have used exhaustive determination.

Edited January 4, 2010 by Medji

[Guest]

I have a theory:

Let each position be named thus: BL B1 B2 B3 B4 = C1 C2 C3 C4

Now, we know that there is an interaction in B2, B3 and B4 that produces C1 by using a sum to determine the relative anti-position of the first element of the code.

My theory is that instead of an interaction in B1, B2, B3 and B4 producing C2, that it is instead an interaction between either B1 and C1, or B1, B2, B3, B4 and C1, with the most likely being an interaction between B1 and C1. I've been evaluating squares, square roots, multiples, divisors, sums, differences, everything I could think of, and I haven't found anything yet. Still working on it, though, until I have used exhaustive determination.

I can see the logic in your theory and agree that it is has potential (I have also pursued a similar line of thought with various calculations...but no success yet !)

Can I just make a couple of additional , general comments.

1....C2 may NOT be the next digit to be decoded. It might by C3 or C4 before C2

2...There may be some form of "carry over" from the calculation for a previous code digit.

Example : if the calculation that produced C1 was a number higher than 10 .... eg 12 ....then the 1 (the "tens" )could possibly be "carried over" and used in a calculation for C2 or C3 or C4 .......This has happened in previous Codebreaker problems and can explain why a theory may work for most of the time, but has a few exceptions.....and these exceptions may be 1 (or 2 etc. ) higher or lower than the result from the main theory. The difference being a "carry over " from the preceding calculation.