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#1 redrooster

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Posted 25 November 2009 - 06:15 PM

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 !
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#2 redrooster

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Posted 25 November 2009 - 06:28 PM

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.
Spoiler for Timesaver......Beware of Rad Herrings ..........

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#3 nutso42

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Posted 26 November 2009 - 03:19 AM

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


Still working on it....
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#4 redrooster

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Posted 29 November 2009 - 12:44 PM

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?
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#5 ljb

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Posted 30 November 2009 - 10:21 PM

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?
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#6 redrooster

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Posted 01 December 2009 - 01:06 PM

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
Spoiler for Hints for first Code Digit

Edited by redrooster, 01 December 2009 - 01:10 PM.

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#7 redrooster

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Posted 01 December 2009 - 06:00 PM

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.
Spoiler for Original Codebreaker Wanted Puzzle AUG 2008

Edited by redrooster, 01 December 2009 - 06:01 PM.

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#8 redrooster

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Posted 04 December 2009 - 11:34 PM

Is there anyone out there taking up this challenge?
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#9 redrooster

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Posted 09 December 2009 - 02:34 PM

Hint /Explanation / Part answer for the first Code Digit
Spoiler for First Code Digit

Edited by redrooster, 09 December 2009 - 02:39 PM.

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#10 redrooster

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Posted 10 December 2009 - 10:49 PM

$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.
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