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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.. 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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Posted · Report post

The 6998 and 6999 has been bugging me, they are consecutive integers with the same C4 of 5. They are close to 7000 where all the box digits will change so perhaps that fact is important. I am also exploring multiple mod functions. For example, (6999 mod 13) mod 5. This could explain the jump in the sequence of C4s in the 020* series.

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The 6998 and 6999 has been bugging me, they are consecutive integers with the same C4 of 5. They are close to 7000 where all the box digits will change so perhaps that fact is important. I am also exploring multiple mod functions. For example, (6999 mod 13) mod 5. This could explain the jump in the sequence of C4s in the 020* series.

It may be worth remembering that the answer for C4 may depend on information contained in an answer for C1 C2 or C3

ie....As C1 has already been calculated, perhaps the next calculation should be C2 , then C3 and finally C4 ( or other variations? )

Edited by redrooster
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is there any logic behind the reason that there is only 1 box containing the letter L and the letter D and the letter W and letter S

L 3865 = 2545

D 3631 = 1152

W 0496 = 2125

S 4705 = 4133

Edited by andy070509
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Posted · Report post

is there any logic behind the reason that there is only 1 box containing the letter L and the letter D and the letter W and letter S

L 3865 = 2545

D 3631 = 1152

W 0496 = 2125

S 4705 = 4133

got this puzzle last week and have been working on it for a bit i have been working in reverse

i.e crack the last digit first and work backwards but i will share when i have working logic solution to my latterall thinking

but are u sure that the letters have no way involved in solving this

i also noticed along with several of u that some boxe's repeat them selves

p.s as u can see i just joined but i have a very good mathematical brain and have won several awards in maths

and i hope that all of us can bounce of each other to get this solved

cheers it was nice to hear some of you thoeries but i have to admit this is a tough 1 to crack

lets hope we do it

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Posted (edited) · Report post

is there any logic behind the reason that there is only 1 box containing the letter L and the letter D and the letter W and letter S

L 3865 = 2545

D 3631 = 1152

W 0496 = 2125

S 4705 = 4133

got this puzzle last week and have been working on it for a bit i have been working in reverse

i.e crack the last digit first and work backwards but i will share when i have working logic solution to my latterall thinking

but are u sure that the letters have no way involved in solving this

i also noticed along with several of u that some boxe's repeat them selves

p.s as u can see i just joined but i have a very good mathematical brain and have won several awards in maths

and i hope that all of us can bounce of each other to get this solved

cheers it was nice to hear some of you thoeries but i have to admit this is a tough 1 to crack

lets hope we do it

Welcome aboard..It`s always good to have a fresh brain on this subject.

I would suggest that you read through the previous messages where you will find that Code Digit 1 (C1) has been solved.

Also there was an original "hint" from the original source that the letters may not be required, and this has been partly verified by Boxes C0209 and Y0209 having the same code 5554 ( This duplicate information is not in my list of C4 digits above (Message 51 ) in order to avoid confusion) so I suspect that your observation of letters L D W S will not be relevant in the solving of the code (but I am willing to be convinced otherwise !)

Edited by redrooster
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Posted · Report post

how come in post 51 the boxes and there codes are different from the ones given at the begginig?

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Posted · Report post

nvm about my last post i didnt look through the whole thing\

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i think the first digit is 2. my reasoning is that all the boxes whos codes start with a 2 the sum of the box numbers last 3 digits are 1 less than 5.

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Welcome to the challenge tiradejr! We've already solved (actually redrooster did I think) the first Code number C1. And you're correct about the first digit also. We're currently working on the last 3 digits, C2 C3 & C4.

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Posted · Report post

are you guys starting with c4 or are you just trying them all?

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am still at it and mostly working on 4 but 2 and 3, too. re 4, am trying to rationalize the jump in the sequence when B4 goes from 7 to 8 in the successive clues last added. converting B4 to base 8 and mod 5 initially looked promising, and also with B2 and B3 when you add B4 back in, but the little time I've been able to give it has produced anything conclusive.

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Ok after trying different mods of B1+B2+B3+B4, mod 8 seems to be fairly consistent with C2. I'm pretty much stumped and don't know where to go from here so I thought I'd share. This is what I have.

(Sum of digits in the box number) mod 8

mod = 0 -> C2 = 1 or 5

mod = 1 -> C2 = 1, 4 or 5

mod = 2 -> C2 = 4 or 5

mod = 3 -> C2 = 4

mod = 4 -> C2 = 3 or 4

mod = 5 -> C2 = 2 or 3

mod = 6 -> C2 = 1, 2 or 3

mod = 7 -> C2 = 1, 2 or 5

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Crap, too late to edit. I'm dumb.

Ok what I meant to say was the mod was of the entire box number not the sum of digits. Which makes more sense now. :duh:

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Posted · Report post

srry this is a dumb question but wat is mod?

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srry this is a dumb question but wat is mod?

A mod is short for modulus which is the remainder after a division. Therefore, 28 mod 5 = 3 because 3 is the remainder after dividing 28 by 5. This is used to determine C1 also.

*edit* I will also mention some examples pertaining to the code.

Box V 1192; Code 4521

1192 mod 8 = 0 and it's C2 is 5

Box J 5264; Code 4524

5264 mod 8 = 0 and it's C2 is 5

Box A 2408; Code 4124

2408 mod 8 = 0 and C2 is 1

Box R 0871; Code 5545

0871 mod 8 = 7 and C2 is 5

Edited by JTZero
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A mod is short for modulus which is the remainder after a division. Therefore, 28 mod 5 = 3 because 3 is the remainder after dividing 28 by 5. This is used to determine C1 also.

*edit* I will also mention some examples pertaining to the code.

Box V 1192; Code 4521

1192 mod 8 = 0 and it's C2 is 5

Box J 5264; Code 4524

5264 mod 8 = 0 and it's C2 is 5

Box A 2408; Code 4124

2408 mod 8 = 0 and C2 is 1

Box R 0871; Code 5545

0871 mod 8 = 7 and C2 is 5

Using the given box number as a 4 digit number with mod 8 is an interesting concept. However I have been involved in quite a few of these codebreaker puzzles before and I wonder whether this is the way to go for this particular problem , especially as C1 was solved using the single digits B2 B3 B4 (but prove me wrong by all means !)

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i think the first digit is 2. my reasoning is that all the boxes whos codes start with a 2 the sum of the box numbers last 3 digits are 1 less than 5.

It might be useful to read the method used on a similar Codebreaker problem set by the same person.

There is a link in the earlier message 07 in this thread.

I am still offering a $50 prize to anyone who manages to crack this.

See details in message 10 earlier in this thread.

Edited by redrooster
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Using the given box number as a 4 digit number with mod 8 is an interesting concept. However I have been involved in quite a few of these codebreaker puzzles before and I wonder whether this is the way to go for this particular problem , especially as C1 was solved using the single digits B2 B3 B4 (but prove me wrong by all means !)

Using this method there was a strange correlation between this mod 8 and C2 as I have mentioned earlier. Although using mod 8 we can eliminate the need for B1 as a factor because B1 can be any value and the mod won't change. That's the nature of multiples of 8. Your note helped me though. I'll be back with some more results.

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Using this method there was a strange correlation between this mod 8 and C2 as I have mentioned earlier. Although using mod 8 we can eliminate the need for B1 as a factor because B1 can be any value and the mod won't change. That's the nature of multiples of 8. Your note helped me though. I'll be back with some more results.

Food for thought?

Message 27 made a useful comment that instead of the remainder being used, consider using the first part , the whole number.

So using mod 5 on 16 the answer is 3 r 1...so instead of using the 1 , use the 3 in any further calculation

Edited by redrooster
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Food for thought?

Message 27 made a useful comment that instead of the remainder being used, consider using the first part , the whole number.

So using mod 5 on 16 the answer is 3 r 1...so instead of using the 1 , use the 3 in any further calculation

That's brilliant! That may even be useful for C4 calculations as well. Thanks!

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Sorry if I'm repeating anything that has already been said ... I didn't look at any of the spoilers because I didn't want someone elses logic starting my thinking proccess before I started thinking ... but I've read all the nonspoilers ... So here's what I'm thinking because I work with inventory. If I had thousands of boxes and four walls to place them this is how I'd do it. Label each wall example: north , south , east, west. Then label each row from A-Z (like the first letter we current don't have a use for) then label each column on each wall. So if i sent someone to go find something I'd say go get my diamonds from the north wall position F,90 or something like that ... therefore there could be 4 different formulas one for each wall and then depening on its position on the wall and which wall not the number on the box(though obviosuly that must help us too) so the code maybe have something to do with it's position on the wall and which wall instead of the label on the box(which might be used to determine the postion of the box and not the code) ... maybe

Edited by PVRoot
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Sorry if I'm repeating anything that has already been said ... I didn't look at any of the spoilers because I didn't want someone elses logic starting my thinking proccess before I started thinking ... but I've read all the nonspoilers ... So here's what I'm thinking because I work with inventory. If I had thousands of boxes and four walls to place them this is how I'd do it. Label each wall example: north , south , east, west. Then label each row from A-Z (like the first letter we current don't have a use for) then label each column on each wall. So if i sent someone to go find something I'd say go get my diamonds from the north wall position F,90 or something like that ... therefore there could be 4 different formulas one for each wall and then depening on its position on the wall and which wall not the number on the box(though obviosuly that must help us too) so the code maybe have something to do with it's position on the wall and which wall instead of the label on the box(which might be used to determine the postion of the box and not the code) ... maybe

I'm sorry after I read it and posted it, it sounds a little arrogant on my part saying "I don't want someone else's logic starting my thinking process" but;

A Wise Man Once Told Me: "Logic can't be taught. It can only be steered"

So this is where I'd start. I still haven't read the spoilers yet I'll do that now. But this is where I'm gonna start. I'll see if I can organize the empty boxes with given code from the original problem on the 4 walls or 3 walls or 2 walls, the problem only states walls, to see if the position somehow determines its code and the label on the box determines its position ... Also if there are all four walls filled from top to bottom the fourth wall has to have a door in it which might cause for gaps in the numbering which I think you guys have noticed but I don't know yet maybe just the three walls and not the door wall yet and the gaps maybe something else.

Edited by PVRoot
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I'm sorry after I read it and posted it, it sounds a little arrogant on my part saying "I don't want someone else's logic starting my thinking process" but;

A Wise Man Once Told Me: "Logic can't be taught. It can only be steered"

So this is where I'd start. I still haven't read the spoilers yet I'll do that now. But this is where I'm gonna start. I'll see if I can organize the empty boxes with given code from the original problem on the 4 walls or 3 walls or 2 walls, the problem only states walls, to see if the position somehow determines its code and the label on the box determines its position ... Also if there are all four walls filled from top to bottom the fourth wall has to have a door in it which might cause for gaps in the numbering which I think you guys have noticed but I don't know yet maybe just the three walls and not the door wall yet and the gaps maybe something else.

An interesting approach ...but after reading the spoilers then I`m fairly sure that it will sway you to think more mathematically....especially if you read through the previous (and similar) puzzle

Edited by redrooster
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Now that I've read the spoilers... I'm thinking I might be thinking to logically and not mathematically(but I'm not giving up on logic or the illogical and I'm too hot and tired to Math today). But I also reread the original problem and a "strong room" could have more then 4 walls. So lets say 5 because if the codes are only 1111-5555 which is 5^4 which is 625 possible combinations(I think) and we are pretty certain that we've solved for C1 using the boxes label(though still that could be coincidental). Then we are only looking for the other 125 possible combinations left. With 5 walls we could have 25 rows and 25 columns which would give us all 625 codes(and positions) times 5. Which would give us 3125 boxes which makes sense because the problem states "thousands of boxes" and give reason for codes repeating because multiple boxes have the same code and there are thousands of boxes and only 625 codes. Now using B1 and B2 and B3 and B4 and possibly the letter we've discarded as useless, to some how come up with 125 different but repeatable outcomes? Instead of trying to find C2 and C3 and C4 separately. Or organize the known boxes on one of the walls(5) with one of the rows(5*5) with one of the columns(5*5) with obvious repeats; AND have it's position on one of the 5 walls with one of the 25 rows with one of the 25 columns somehow determine it's code. But I don't know how to do that either.

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Ok I am posting to show that I am still working on this. I have decided to split the boxes into odds and evens to tackle C2 and C3. C4 seems fairly linear and is comparable to C1. Also decided to try squares of digits instead of multiples.

Also, to redrooster, did anyone finish the first codebreaker problem?

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