[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]

Here are my thoughts.

We know that all numbers have to be between one and five. The only mathematical way to consistently get a number between one and five... that I know of is with modulo function. Modulo is simply the remainder after division. In c++ (programing) the '%' symbol is used to represent modulo function so I will use that for my example.

The best formula I can come up with for the first number is

C1 = 5 - ((D2 + D3 + D4 - 1 ) % 5)

Their are a few other ways to calculate this but the important part is that they all involve the remainder.

Now let's have a look at whats going on.

Z 4725 = 2325

7 + 2 + 5 = 14

14 - 1 = 13

13 / 5 = 2 r 3

5 - 3 = 2

C1 = 2

When we divide 13 by 5 we get 2 remainder 3 but only the remainder is used for the calculation of C1. I'm thinking that we must use the whole number in the next step of the puzzle.

Another way to look at it is to convert the number to base five.

eg:

0 = 0

1 = 1

2 = 2

3 = 3

4 = 4

5 = 10

6 = 11

7 = 12

8 = 13

9 = 14

10 = 20

11 = 21

12 = 22

13 = 23

14 = 24

15 = 30

......

7 + 2 + 5 = 14

14 = 24 (in base five)

24-1=23

5 - 3 = 2

This is just another way of looking at the first formula I showed.

the reason we need to subtract 1 is to convert from regular base 5 ( 0 - 4) to the riddles base 5 ( 1 - 5)

I will keep looking to see if I can find anything else but this might help get someone on the right track.

[Guest]

Here are my thoughts.

We know that all numbers have to be between one and five. The only mathematical way to consistently get a number between one and five... that I know of is with modulo function. Modulo is simply the remainder after division. In c++ (programing) the '%' symbol is used to represent modulo function so I will use that for my example.

The best formula I can come up with for the first number is

C1 = 5 - ((D2 + D3 + D4 - 1 ) % 5)

Their are a few other ways to calculate this but the important part is that they all involve the remainder.

Now let's have a look at whats going on.

Z 4725 = 2325

7 + 2 + 5 = 14

14 - 1 = 13

13 / 5 = 2 r 3

5 - 3 = 2

C1 = 2

When we divide 13 by 5 we get 2 remainder 3 but only the remainder is used for the calculation of C1. I'm thinking that we must use the whole number in the next step of the puzzle.

Another way to look at it is to convert the number to base five.

eg:

0 = 0

1 = 1

2 = 2

3 = 3

4 = 4

5 = 10

6 = 11

7 = 12

8 = 13

9 = 14

10 = 20

11 = 21

12 = 22

13 = 23

14 = 24

15 = 30

......

7 + 2 + 5 = 14

14 = 24 (in base five)

24-1=23

5 - 3 = 2

This is just another way of looking at the first formula I showed.

the reason we need to subtract 1 is to convert from regular base 5 ( 0 - 4) to the riddles base 5 ( 1 - 5)

I will keep looking to see if I can find anything else but this might help get someone on the right track.

Welcome back toxicoxyde..I was hoping that you would take a look at this problem. It is a lot harder than that last one in Aug 2008....and I found some of that rather difficult at the time ...or at least until someone pointed out the multiple of 7 idea and the remainder solution to some of the code digits.

I can follow all your reasoning ( and it is interesting that a formula can be created to solve C1 )

I have now spent a fair bit of time on this problem and just cannot see a common point for C2 C3 or C4.

As I pointed out earlier, I am convinced that B1+B2+B3+B4 are all involved in the next stage......but what that is has not been realised yet.

Is anyone else working on any theories that they would like to share?

Perhaps we ought to have a " brainstorming" session for a bit of lateral thinking.

[plainglazed]

I hesitate to mention this cause it's gotten me nowhere but anyway: Whenever the box number ends in 5, the code also ends in 5 in all but one instance where it is 3. The probability of such a distribution is very slim. In the instance where the last box digit is 5 and the last code digit is not 5 but 3, the third box digit is zero and the third box digit in every other case is not zero. Since it is likely most code digits are determined by converting some calculation mod 5, I thought that maybe the last code digit might have something to do with the fourth box digit raised to the power of the third box digit. There is also a somewhat odd distribution of last code digits when the last box digit is 1.

[Guest]

I hesitate to mention this cause it's gotten me nowhere but anyway: Whenever the box number ends in 5, the code also ends in 5 in all but one instance where it is 3. The probability of such a distribution is very slim. In the instance where the last box digit is 5 and the last code digit is not 5 but 3, the third box digit is zero and the third box digit in every other case is not zero. Since it is likely most code digits are determined by converting some calculation mod 5, I thought that maybe the last code digit might have something to do with the fourth box digit raised to the power of the third box digit. There is also a somewhat odd distribution of last code digits when the last box digit is 1.

An interesting observation.

All we need now is to see how they are related mathematically.

[Guest]

Naming the positions as such : CL C1 C2 C3 C4 = P1 P2 P3 P4

Observation 1

For the following, (C2 + C3 + C3 + C4) (mod 5) = (P4)

A 9024=5153

E 1390=4251

V 1192=4521

Z 2494=4221

K 1533=5244

Z 4108=2414

Y 0926=4254

K 9356=2424

V 1044=3442

Z 2271=1552

E 8391=3232

Z 2812=5332

A 7713=5542

A 8930=4545

H 1381=4255

R 0464=2135

A 4089=4455

L 3865=2545

Observation 2

When C1+C2 = 4, then P2 = Cx1 or Cx2.

(Using the conversion of C1 mod(5) or C2 mod(5) (like solving for the the first digit, P1) with the following):

C1 to Cx1

5=1

1=2

3=3

4=4

2=5

C2 to Cx2

4=1

3=2

1=3

5=4

2=5

Ex:

R 0448=5132

(0+4) = 4, C1 mod (5) = 5, 5 converts to 1 using the Cx1 chart, which = P2

or (0+4) = 4, C2 mod (5) = 4, 4 converts to 1 using the Cx2 chart, which = P2

E 1390=4251

(1+3) = 4, C1 mod (5) = 1, 1 converts to 2 using the Cx1 chart, which = P2

or (1+3) = 4, C2 mod (5) = 3, 3 converts to 2 using the Cx2 chart, which = P2

and works for these as well:

Z 2271=1552

H 1381=4255

R 0464=2135

A 4089=4455

W 0496=2125

I haven't seen this combination for any other sums of C1+C2...

Something obvious might be sticking out but I can't see how these two observations apply to the whole population.

Edited January 29, 2010 by Matthew

[Guest]

A couple of interesting observations Matthew

To save any confusion , can we use the shorthand of B = Box and C = Code for the digits

BL B1 B2 B3 B4 = C1 C2 C3 C4

Matthew .....I have taken the liberty of rewriting your spoiler using the B and C.

Two Observations (rewritten)

Observation 1

For the following, (B2 + B3 + B3 + B4) (mod 5) = (C4)

A 9024=5153

E 1390=4251

V 1192=4521

Z 2494=4221

K 1533=5244

Z 4108=2414

Y 0926=4254

K 9356=2424

V 1044=3442

Z 2271=1552

E 8391=3232

Z 2812=5332

A 7713=5542

A 8930=4545

H 1381=4255

R 0464=2135

A 4089=4455

L 3865=2545

Observation 2

When B1+B2 = 4, then C2 = Bx1 or Bx2.

(Using the conversion of B1 mod(5) or B2 mod(5) (like solving for the the first digit, C1) with the following):

B1 to Bx1

5=1

1=2

3=3

4=4

2=5

B2 to Bx2

4=1

3=2

1=3

5=4

2=5

Examples :

R 0448=5132

(0+4) = 4, B1 mod (5) = 5, 5 converts to 1 using the Bx1 chart, which = C2

or (0+4) = 4, B2 mod (5) = 4, 4 converts to 1 using the Bx2 chart, which = C2

E 1390=4251

(1+3) = 4, B1 mod (5) = 1, 1 converts to 2 using the Bx1 chart, which = C2

or (1+3) = 4, B2 mod (5) = 3, 3 converts to 2 using the Bx2 chart, which = C2

and works for these as well:

Z 2271=1552

H 1381=4255

R 0464=2135

A 4089=4455

W 0496=2125

I haven't seen this combination for any other sums of B1+B2...

Something obvious might be sticking out but I can't see how these two observations apply to the whole population.

Edited January 31, 2010 by redrooster

[Guest]

Naming the positions as such : CL C1 C2 C3 C4 = P1 P2 P3 P4

Observation 1

For the following, (C2 + C3 + C3 + C4) (mod 5) = (P4)

A 9024=5153

E 1390=4251

V 1192=4521

Z 2494=4221

K 1533=5244

Z 4108=2414

Y 0926=4254

K 9356=2424

V 1044=3442

Z 2271=1552

E 8391=3232

Z 2812=5332

A 7713=5542

A 8930=4545

H 1381=4255

R 0464=2135

A 4089=4455

L 3865=2545

Observation 2

When C1+C2 = 4, then P2 = Cx1 or Cx2.

(Using the conversion of C1 mod(5) or C2 mod(5) (like solving for the the first digit, P1) with the following):

C1 to Cx1

5=1

1=2

3=3

4=4

2=5

C2 to Cx2

4=1

3=2

1=3

5=4

2=5

Ex:

R 0448=5132

(0+4) = 4, C1 mod (5) = 5, 5 converts to 1 using the Cx1 chart, which = P2

or (0+4) = 4, C2 mod (5) = 4, 4 converts to 1 using the Cx2 chart, which = P2

E 1390=4251

(1+3) = 4, C1 mod (5) = 1, 1 converts to 2 using the Cx1 chart, which = P2

or (1+3) = 4, C2 mod (5) = 3, 3 converts to 2 using the Cx2 chart, which = P2

and works for these as well:

Z 2271=1552

H 1381=4255

R 0464=2135

A 4089=4455

W 0496=2125

I haven't seen this combination for any other sums of C1+C2...

Something obvious might be sticking out but I can't see how these two observations apply to the whole population.

Just a bit of "open thinking " re the two observations

Just thinking about the two observations from the previous answer from Matthew

Observation 1

For the following, (B2 + B3 + B3 + B4) (mod 5) = (C4)

This is an interesting concept ..doubling the value of one of the digits in any calculation

.However, I believe that care has to be taken before jumping to any conclusions with any theory due to the fact that only 5 digits are used in the Code answer ( digits 1 2 3 4 5 ) so whatever theory is considered , there is a 4 - 1 chance that it will appear to provide the correct answer.

Observations such as if B4 = 5 then C4 = 5 as shown below appear possible

E 0215 = 3235

R 0545 = 2545

U 1095 = 2245

Z 4725 = 2325

A 7535 = 3555

...but is disproved because of the following ending in 3

S 4705 = 4133

Observation 2

When B1+B2 = 4, then C2 = Bx1 or Bx2.

(Using the conversion of B1 mod(5) or B2 mod(5) (like solving for the the first digit, C1) with the following):

This one is more interesting as it appears to work in all examples listed where B1+ B2 = 4

However extending the logic , if B1 + B2 = the same number then C2 should give the same answer.

It does in

E 0215 = 3235

H 0262 = 1234

R 0448 = 5132

R 0464 = 2135

but not in

A 0859 = 4453

R 0871 = 5545

Both suggestions offer a few more ideas though

Edited January 31, 2010 by redrooster

[Guest]

Update: I don`t know about anyone else but I`m having no success working out C2 C3 or C4

I have been in contact with the originator via one of the students and have been told that only ONE student in the 2008 class managed to crack this codebreaking problem.(..and he did NOT divulge the answer or give any clues to the method used )

The same problem was then set for the following year`s students , only this time there were extra Box Numbers with Codes given.....so I am in the process of obtaining this extra list of Box/Code information.

Hopefully this might provide us with relevant (and useful ) extra information .

I am still offering a $50 USD prize as mentioned in an earlier message and as I am that keen to find a solution I will add an extra $10 USD for the person who solves the method used for the final Code digit (C4 ?)

I will post the extra information as soon as I get it.(within the next few days).

[Guest]

Does anyone think that it could be even harder than what you've found so far? Meaning the letters are needed as well to find an equation for the digits?

[Guest]

Does anyone think that it could be even harder than what you've found so far? Meaning the letters are needed as well to find an equation for the digits?

That is always possible.

The only reason I mentioned in the beginning that perhaps the letters were not used was because , when asking the original creator of this Codebreaker problem for any possible clues (as it is a very difficult problem!) , he said that perhaps not all of the Box Number information quoted was used in any code calculation solution .

The general consensus by the students (who were working on this problem at the time ) took this to mean that it was the letter part that was not needed as several of the students (who claimed to have solved more than just C1 as we have ) said that they had used all four box digits (B1 B2 B3 B4 ) when working on their answers to other parts of the code.

Perhaps this is not correct?

However as stated in my previous message only ONE student from 2008 had managed to crack the complete method used , so extra information has been given in order for the next batch of students (2009 group ) to help provide a few further clues.....and hopefully the solution

I am still chasing up this extra page of Box Number/ Code information and will post it here as soon as I receive it.

[jazzship]

Same thing, but just out interest for C1 I got:

(B2 + B3 + B4 + C1) mod 5 = 1

I haven't been too systematic but it seems like I've tried every obvious arithmetic calculation (similar to solution for C1) with two, three and four numbers including C1 to get C2, so far unsuccessfully. I got a feeling it required a systematic progression involving C1 but I haven't got anywhere and now I think its likely C2 requires a technique somewhat different to the technique for solving C1, though I may have missed something obvious. It's hard to imagine that Mod 5 doesn't make another appearance though.

Intriguing that the letters might have been added for no other reason than to confuse. Very interested to see the answer if someone can solve this.

Edited February 22, 2010 by jazzship

[Guest]

Very interested to see the answer if someone can solve this.

So am I !!

I have been trying to crack this delightfully difficult codebreaking problem for around 6 months now and have only managed to solve the C1.

Hence why I`ve offered a reward !

I`ve been promised the further information on Tuesday ,so fingers crossed, there might be a few more clues to help us solve it..

[Guest]

As promised , here is the extra page of further clues .

Let`s hope they prove useful

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

[Guest]

There were three more clues missing from the last list.

Here they are

K 9156 = 4323

J 0626 = 2512

C 3021 = 3221

[Guest]

Is anyone still working on this Codebreaker problem?

If so , then I `ve been told that it has been comfirmed that the initial LETTER is NOT used in any calculation.(This was hinted at in the initial posting )

ie...the code will be dependent on just the box number digits ... B1 B2 B3 and B4 only

This means that the answer will be purely mathematical !

Unfortunately , I have not been given any further clues.

Has anyone got any brainstorming ideas that they are working on?

I am going to start afresh and see if I`ve missed anything.

It IS solvable , so it makes a very interesting , but rather difficult problem to crack.

[Guest]

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

[Guest]

I can't say that I'm "still" working on this - I just found it because of your most recent post (I'm rather new to the board) - I'm gonna take a crack at it though - maybe some fresh eyes. I'll work on it for awhile (in days/weeks/months) though. Thanks for posting this - I think

[Guest]

My Initial two observations:

In the data given - the same box number appears only once - C 0209 & Y 0209.

The code for both of these is the same - so the conclusion that box letter can be disregarded is somewhat verifiable

In the given data - 11 codes are repeated - only once is the box number the same.

R 0545 2545

L 3865 2545

R 0040 2551

C 0202 2551

E 4111 3222

A 7238 3222

C 0206 3245

A 6053 3245

C 0201 3555

A 7535 3555

Z 1043 4455

A 4089 4455

A 4953 4531

E 8250 4531

R 1335 5153

A 9024 5153

J 1853 5322

K 2380 5322

Z 1795 5452

Z 5970 5452

C 0209 5554

Y 0209 5554

I haven't drawn any conclusions about the second observation - but I wanted to get it out there.

[Guest]

I can't say that I'm "still" working on this - I just found it because of your most recent post (I'm rather new to the board) - I'm gonna take a crack at it though - maybe some fresh eyes. I'll work on it for awhile (in days/weeks/months) though. Thanks for posting this - I think

Welcome aboard ! Let`s hope a new enthusiast will give us some original ideas deas and open up a few new avenues.

Any brainstorming theories or observations are always most welcome, no matter how obscure .

I am already aware of the observations that you have mentioned so far ...but keep them coming !

As you may realise , the first Code Digit C1 has been solved ....and so now I`m stuck.

note I am still offering a $50 USD prize for solving this codebreaking problem as I am really intrigued to find the working solution.

[Guest]

could you post the source of the problem?

[Guest]

could you post the source of the problem?

Originally set as one of several mind stimulating "homework/vacation " projects by the Maths Lecturer on my son`s Engineering Course at University.

For confidentiality reasons, I would prefer not to name the exact University.

As an avid crossword and mathematical problem enthusiast, I was asked for advice on a previous (similar) problem ....and I became hooked .

Over the past few years , there have been many "mind stimulating" problems set for vacation homework ...but this one is the hardest so far and I`ve been told that it has only ever been solved by one person.....and at the moment , it looks as though it will remain like that...(unless someone can discover an answer to the next stage.( C2 perhaps?)

Any suggestions or theories most welcome ...or has everyone on here now given up?

[Guest]

Any suggestions or theories most welcome ...or has everyone on here now given up?

I don't get to crack at it a lot but I'm actually working on C4 - you already pointed out the relationship between B4 and C4 in the 020# series. I'm sure you notice the deviation that occurs at 0208 - since the quotient of (B2+B4),10 is 1 - B4 jumps by 2 instead of 1 like from 0206 to 0207.

I'm working on the actual algorithm for this relationship as well as how B1, B2 & B3 play into it - several theories but none fleshed out well enough to pass on. I got close once but couldn't quite make it work for all cases.

This is very challenging - I like it very much.

Thanks for posting.

[Guest]

Hey guys, I'm new here and ready to take on some challenges! This codebreaker puzzle caught my attention and I decided to join the forums. I hope I can be of some help in solving this one.

[Guest]

Those extra boxes kinda helped. Here's an observation I found for C4.

For these pairs of consecutive boxes, the C4s repeat the pattern 5 2 in each pair. The two pairs are 52 box numbers apart and 52 happens to be a multiple of 13 (nice prime number!). From the series of C0200 to C0209 we see the sequence continues for at least 8 consecutive boxes. Between 0207 and 0208, C4 is either being increased by 2 or decreased by 3. 0208 is divisible by 13 also. 1040 and 1092 are divisible by 13 and each are equally close to these pairs but there is a difference from the C020* sequence. Instead of the -3 gap (that's what I believe it is) occurring near this multiple of 13, it's farther away. I also noticed in my two pairs of consecutive boxes, (B1 + B2? + B3) mod 5 = 4 whereas in the 020* sequence it's 2. It's possible that C4 requires these other values but the key number I have found for solving C4 is 13.

Z1043 = 4455

V1044 = 3442

&

U1095 = 2245

U1096 = 1142

Also, check the box numbers around 4*** for close numbers to test (currently checking them now).

[Guest]

As there seems to be interest in attempting to solve C4 next , then here is the full list of Box/Code digits arranged in the order of the same C4 digit

R 0040 = 2551

H 0124 = 4441

C 0202 = 2551

C 0207 = 2231

V 1192 = 4521

E 1390 = 4251

Z 2494 = 4221

Z 3229 = 3321

A 3643 = 3411

A 4953 = 4531

A 5209 = 5121

K 6620 = 3341

A 7962 = 4431

E 8250 = 4531

C 0203 = 1442

R 0448 = 5132

A 0704 = 5122

V 1044 = 3442

C 1071 = 3252

U 1096 = 1142

Q 1167 = 2232

Z 1795 = 5452

J 1853 = 5322

Z 2271 = 1552

K 2380 = 5322

Z 2812 = 5332

D 3631 = 1152

E 4111 = 3222

A 4163 = 1452

K 4431 = 3552

Z 5970 = 5452

A 7238 = 3222

A 7713 = 5542

E 8108 = 2332

E 8121 = 2522

E 8391 = 3232

A 8844 = 5432

A 9751 = 3242

C 0138 = 4513

C 0204 = 5423

C 0208 = 1133

A 0859 = 4453

R 1335 = 5153

R 1479 = 1523

A 2593 = 4553

A 4476 = 4343

S 4705 = 4133

E 5567 = 3123

E 5862 = 5223

K 6481 = 3513

U 8056 = 5543

A 9024 = 5153

K 9139 = 3443

J 9774 = 3253

C 0200 = 4134

C 0205 = 4314

Y 0209 = 5554

C 0209 = 5554

H 0262 = 1234

Z 0869 = 3324

Y 0926 = 4254

K 1533 = 5244

A 2356 = 2324

C 3010 = 5414

Z 4108 = 2414

R 5264 = 4524

E 7402 = 5454

Z 8101 = 4214

K 9356 = 2424

A 9396 = 3354

Z 9800 = 3134

J 0023 = 1215

A 0158 = 2255

C 0201 = 3555

C 0206 = 3245

E 0215 = 3235

R 0464 = 2135

W 0496 = 2125

R 0545 = 2545

E 0843 = 1455

R 0871 = 5545

Z 1043 = 4455

U 1095 = 2245

R 1244 = 1355

H 1381 = 4255

E 2182 = 5225

A 2994 = 4425

E 3695 = 1545

L 3865 = 2545

A 4089 = 4455

A 4166 = 3325

A 4707 = 2455

Z 4725 = 2325

Z 5271 = 1225

J 5362 = 5535

A 5454 = 3255

A 6053 = 3245

Z 6163 = 1445

J 6242 = 3545

A 6322 = 4525

J 6519 = 1525

J 6709 = 5355

J 6998 = 5235

E 6999 = 4515

A 7535 = 3555

J 8102 = 3155

Z 8596 = 1335

A 8930 = 4545

E 9153 = 2515