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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

Also , it is 99% certain that the first letter is not relevant ...so is no part of the solution

**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 **5**271 = 1**225**

Z **2**271 = 1**552**

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

V 1**0**44 = **3442**

R 1**2**44 = **1335**

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

S 47**0**5 = **4133**

S 47**2**5 = **2325 **

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

S 470**5** = **4133**

A 470**7** = **2455**

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

Z 104**3** = **4455**

V 104**4** = **3442**

U 109**5** = **2245**

U 109**6** = **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**

**redrooster**- Replies: 2

I enjoy decoding challenges and 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 either with suggestions or answers?

This was set as a holiday project for University Engineering Students and as far as I know , nobody managed to solve it !

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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 are that the letters may **not** be involved in the calculations (but I`m not 100% sure ) and also each digit of the code may be solved separately. ..so there might be four separate methods to calculate each digit of the code (but they may be interlinked).

Good luck !

redrooster

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