redrooster

EXTRA CLUES Is anyone still working on this difficult Codebreaker problem ? If so , then you may be interested in the following. The original compiler of this problem has admitted that this is a difficult one to solve , so he has given a few extra clues for the series of Boxes C0200 > C 0209 Code Digit FOUR (C4) looks interesting. C 0200 = 4134 C 0201 = 3555 C 0202 = 2551 C 0203 = 1442 C 0204 = 5423 C 0205 = 4314 C 0206 = 3245 C 0207 = 2231 C 0208 = 1133 C 0209 = 5554

I don`t know whether anyone is still working on this Codebreaking problem. I am having absolutely no success in working it out. This was set as a project for a group of Engineering Students in 2008 ..and only one student managed to crack it. (The method used was not divulged) The same problem was set again in 2009 for another group of students , and to make it easier , the creator added a further list of clues to the original list. I`ve managed to obtain a copy of this second list , so if anyone is interested then here it is. Let me know if anyone has any suggestions. BOX........CODE R 0040 = 2551 A 0158 = 2255 E 0843 = 1455 Z 0869 = 3324 C 1071 = 3252 Q 1167 = 2232 R 1335 = 5153 Z 1795 = 5452 A 2356 = 2324 A 2593 = 4553 A 3643 = 3411 E 3695 = 1545 A 4163 = 1452 A 4953 = 4531 A 5209 = 5121 A 5454 = 3255 E 5862 = 5223 J 6998 = 5235 Z 8101 = 4214 J 8102 = 3155 E 9153 = 2515 A 9396 = 3354 J 9774 = 3253

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

Here is the method used to discover how to calculate the FIRST code digit [hide] Method used After making a list of all the Codes that began with the SAME FIRST Code digit it was noticed that adding the 2nd + 3rd + 4th Box Number digits together gave totals that were related. ie...If a multiple of 5 was subtracted from this total , the answer always gave the same final digit.(but this was NOT the first Code Digit) Example :Using a FIRST Code digit of 4 H0124 = 4441 Adding 1+2+4 = 7 minus (1x5) =2 C0138 = 4513 Adding 1+3+8 = 12 minus (2x5) = 2 Z2494 = 4221 Adding 4+9+4 = 17 minus (3x5) = 2 A2994 = 4425 Adding 9+9+4 = 22 minus (4x5) = 2 E6999 = 4515 Adding 9+9+9 = 27 minus (5x5) = 2 So it can be seen that if the initial total is 7 or 12 or 17 or 22 or 27 the final total (after subtracting a multiple of 5) is always 2 .....and this always corresponds to a FIRST Code Digit of 4 (Remember that the Code Number consists of numbers 1, 2, 3, 4, and 5 so this is why the "subtraction of multiples of 5 " is relevant . Similarly applying the same method to other Box Numbers/Code Numbers with different FIRST code digits of 1, 2, 3 and 5, from the list produces the following table If the final total = 1 the FIRST Code digit is 5 if 2 = 4 if 3 = 3 if 4 = 2 if 0 = 1 [/hide]

Not really bizarre . The question requires you to open Box J6383 which has the list inside.(Remember this list contains the identities of the boxes containing the most valuable diamonds ) So you have to work out + know the method of calculation of the code as well because you will be opening other boxes using this list. Otherwise , if the problem was just to open one box (to obtain a piece of paper ) you could just sit there for a few hours trying all 625 different combinations (5 x 5 x 5 x 5 ) on Box J6383 until it opened . Presumably the time factor is important in this diamond theft , hence it is more logical to work out the method of calculation of these codes for maximum gain of opening as many boxes as possible within the allotted time. --------------------------------------------------------------------- I have managed to work out the method used for calculating the FIRST digit of the Code Number...so I am now (still) trying to work out the method for the 2nd, 3rd & 4th digits I intend to post the first digit answer here later , if required ...but was leaving it for the moment to allow forum members time to attempt it themselves. If you want any hints or the answer via a PM then please let me know. Merged post follows: Consecutive posts merged Please post the exact method you would use to find the answer...and we`ll see if it works.

Sorry..the above theory does not appear to work

This is one of the logical starting points where the digits of both the Box numbers and the Code numbers can be compared in order to deduce any connection. Hint ? As we are told that the Code number consists of digits no higher than 5.....so a useful hint is to think about what could be done with any combination of digits that exceed 5.

Yes..I already have many sheets are squared paper with various arrangements of the digits. This has lead me to deduce the method for the first code digit calculation....so there`s only 3 more to go! I was hoping that a few new/fresh brains on this forum might offer me a few more different ideas ...or better still , answers

I have been attempting this very difficult codebreaker problem for the past few weeks and can get no further than finding the answer for just the first digit of the code. Can anyone on here help? This was set as a holiday project for University Engineering Students and as far as I know , nobody managed to solve it ! ========================================================== 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.. Many ( but not all ) of the storage boxes 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 =========================================================================== The only hints I can offer is that the letters may not be involved in the calculations (but I`m not 100% sure ) and also each digit of the code is solved separately. ..so there will be four separate methods to calculate each digit of the code. Good luck ! redrooster