[Guest]

Here is a code breaking problem set by my son`s Engineering Maths Lecturer as “something to keep your mind active” during the Summer Vacation…..and he can only solve half of it so far.

Can anyone help?

===========================================================

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 is several million of ££s .

Each box is labelled with a letter and three digits and the thief knows that the combination lock code for each individual box can be calculated from this sequence.

Eg ...........Box No T176 has a Combination Lock Code of 6896

However , each box has this 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.

The aim is to open two particular boxes which contain the most valuable bag of diamonds

So the question is .

"What is the code for boxes W456 and R588 ?"

Here is the list of open BOX numbers with their security CODES from inside.

Can you workout ( and explain ! ) the method to determine the combination lock code of Boxes W456 and R588

BOX CODE BOX CODE BOX CODE

P194 0851 P327 0606 P541 0427

Q068 9969 Q409 9511 Q638 9382

R002 8940 R012 8947 R017 8990

R025 8956 R030 8900 R031 8970

R032 8941 R033 8912 R037 8994

R054 8986 R082 8946 R111 8847

R125 8827 R142 8819 R150 8874

R163 8883 R216 8761 R228 8700

R283 8758 R291 8714 R294 8726

R342 8650 R370 8610 R395 8667

R437 8577 R544 8430 R560 8454

R568 8410 R642 8362 R679 8358

R711 8261 R754 8271 R808 8120

R836 8189 R846 8186 R854 8142

R870 8163 R876 8187 R941 8004

S526 7477 S702 7232 S982 7079

T337 6601 T668 6384 T860 6160

V120 4871 V555 4409 V950 4038

X482 2521 X613 2334 X623 2331

[Guest]

Brainstorming suggestion?

As we are now looking for the fourth digit of the code , try using this list (where the this fourth digit of the code is the same .... 7)

Assuming that the numbers of these boxes should combine together to eventually produce this same fourth digit of 7 (but not necessarily straightaway)

What connection can be made with the box number (on the left and) and the fourth digit

of the codes (shown on the right).....

R125.....8827

R437.....8577

S526.....7477

R012.....8947

R876.....8187

P541.....0427

R395.....8667

Similar to the third digit calculation , I anticipate that it will be some form of arithmetic

using some/all of the numbers on the left producing a result that will be the same figure

(not necessarily the final answer at this stage , which in this case is 7 .This can be deduced later)

Any suggestions?

The letter directly affects the calculation of the last digit, how is what we need to figure out. That's why I think the consecutive R series might be intentionally placed there as a guide.

if you look at the list I made the only variable that's different in each box# is the last digit and that directly affects the last digit of the code.

R030 8900

R031 8970

R032 8941

R033 8912

R037 8994

oddly enough the first 2 both render a 0 outcome....why?

R030 8900

R031 8970

The only thing i can think of, is that the product of either the last 2 or all 3 of the numbers comes out to 3, thus on some kind of chart as before 3 equates to a 0 on the list, but this is faulty as well because

R941...8004

P194...0851

should be the same 9x4x1...1x9x4

Also if the last 2 digits are the same in any box# the last digit of the code doesn't necessarily stay the same.

e.g.

R082...8946

X482...2521

So we must assume that the letter (thus the first digit) influences the outcome of the 4th row. Any Ideas?

[Guest]

Perhaps the answer is within the 3 digits of the code we've calculated already... just an idea.

[Guest]

oddly enough the first 2 both render a 0 outcome....why?

R030 8900

R031 8970

The only thing i can think of, is that the product of either the last 2 or all 3 of the numbers comes out to 3, thus on some kind of chart as before 3 equates to a 0 on the list,

A couple of comments re the above which might help in the “brainstorming” of this problem

(i)re The product statement above re last 2 or all 3 when multiplying by 0 (zero) will be 0 and not 3 as stated

(ii) re ….R030 8900

R031 8970

both having a 4th code digit answer of 0. ......ie. The same answer appears more than once in a sequence

I can think of a couple of sequences where this may be true

a)….I think it`s called a Fibonacci Sequence where the following digit is the sum of itself plus the previous.

1 , 1 , 2 , 3 , 5 , 8 , 13, 21, 34, 55 , 89 etc.

This sequence could be utilised in a code by just using the last digit giving a sequence of

1 , 1 , 2, 3, 5, 8, 3, 1, 4, 5, 9,

ie the same answer digit 1 can be from three different positions from the sequence ( Positions 1 , 2 and 8 )

b) The other sequence could be if a multiple of 8 (or other even number) was used….

Multiple of 8 gives

8, 16, 24,32,40,48,56,64,72,80

and using just the last digit , the sequence gives

8,6,4,2,0 8,6,4,2,0

where the answer 8 can be found in two positions of the sequence (similarly 6,4,2 and 0)

However this won`t work in our case because the answer is always an even number

[Guest]

HELP !!!!

I`m still looking for ideas for the method to find the final FOURTH digit of the code.

Any suggestions?

I`m convinced that if there is a zero in either the listed BOX or CODE then this might help

work out the final FOURTH digit of the code.

I`ll be posting BOX/CODE Lists with a 0 ZERO in different places and

hope someone can see where the answer might be....or at least offer suggestions

Why 0 ?(zero)

With a 0 (zero) in either the BOX or CODE , then it can be sometimes

easier to spot a pattern as ADDING a 0 can be spotted as making no difference

and also multiplying by 0 always = 0

Eg...When we were looking for the THIRD code digit it can be seen from this list below

that whenever BOX digit one and BOX digit three were added together

they came to the same figure.

However looking at Box R031 it adds up to 7 giving a starting point

clue to where to start deducing the answer. (see Spoiler for answer )

0 in first digit position of BOX

BOX..........CODE

Q068........9969

R002........8940

R012........8947

R017........8990

R025........8956

R030........8900

R031........8970

R032........8941

R033........8912

R037........8994

R054........8986

R082........8946

Further lists to follow , with the 0 in all

positions of both BOX and CODE

Adding BOX digit one + BOX digit three ...then multiplying

the units part of the total by seven...

.and the THIRD code digit figure is the units part of this total.

This means that for any third digit of the code ,

the total of the two digits added together,

.... using the units digit multiplied by 7 ...and

then using the units digit in the total will give a

final third code digit as follows.

eg R032 = 0 + 2 = 2 x 7 = 14 Answer is

for the third code digit is 4

0 x 7 = 10.....Third Code digit is 0

1 x 7 = 7.......................... 7

2 x 7 = 14.........................4

3 x 7 = 21.........................1

4 x 7 = 28.........................8

5 x 7 = 35 ........................5

6 x 7 = 42.........................2

7 x 7 = 49 ........................9

8 x 7 = 56.........................6

9 x 7 = 63.........................3

[Guest]

HELP !!!!

I`m still looking for ideas for the method to find the final FOURTH digit of the code.

Any suggestions?

I`m convinced that if there is a zero in either the listed BOX or CODE then this might help

work out the final FOURTH digit of the code.

I`ll be posting BOX/CODE Lists with a 0 ZERO in different places and

hope someone can see where the answer might be....or at least offer suggestions

Why 0 ?(zero)

With a 0 (zero) in either the BOX or CODE , then it can be sometimes

easier to spot a pattern as ADDING a 0 can be spotted as making no difference

and also multiplying by 0 always = 0

Eg...When we were looking for the THIRD code digit it can be seen from this list below

that whenever BOX digit one and BOX digit three were added together

they came to the same figure.

However looking at Box R031 it adds up to 7 giving a starting point

clue to where to start deducing the answer. (see Spoiler for answer )

0 in first digit position of BOX

BOX..........CODE

Q068........9969

R002........8940

R012........8947

R017........8990

R025........8956

R030........8900

R031........8970

R032........8941

R033........8912

R037........8994

R054........8986

R082........8946

Further lists to follow , with the 0 in all

positions of both BOX and CODE

Adding BOX digit one + BOX digit three ...then multiplying

the units part of the total by seven...

.and the THIRD code digit figure is the units part of this total.

This means that for any third digit of the code ,

the total of the two digits added together,

.... using the units digit multiplied by 7 ...and

then using the units digit in the total will give a

final third code digit as follows.

eg R032 = 0 + 2 = 2 x 7 = 14 Answer is

for the third code digit is 4

0 x 7 = 10.....Third Code digit is 0

1 x 7 = 7.......................... 7

2 x 7 = 14.........................4

3 x 7 = 21.........................1

4 x 7 = 28.........................8

5 x 7 = 35 ........................5

6 x 7 = 42.........................2

7 x 7 = 49 ........................9

8 x 7 = 56.........................6

9 x 7 = 63.........................3

0 (zero) in second digit position of BOX

BOX...........CODE

Q409........9511

R002........8940

R808........8120

S702........7232

0 (zero) in third digit position of BOX

R030........8900

R150........8874

R370........8610

R560........8454

R870........8163

T860........6160

V120........4871

V950........4038

0 (zero) in first digit position of CODE

P194........0851

P327........0606

P541........0427

0 (zero) in second digit position of CODE

R941........8004

S982........7079

V950........4038

0 (zero) in third digit position of CODE

P327........0606

R030........8900

R228........8700

R941........8004

T337........6601

V555........4409

0 (zero) in fourth digit position of CODE

R002........8940

R017........8990

R030........8900

R031........8970

R228........8700

R342........8650

R370........8610

R568........8410

R808........8120

T860........6160

[Guest]

Help still required on this very last stage....the fourth code digit.?

Is anyone looking into this one?

If so , please post your thoughts.

[Guest]

An observation.

R002.....8940

R012.....8947

R032.....8941

R082.....8946

Previously , the third code digit was calculated using a sum of two numbers that was then multiplied by 7

So thinking that multiplying by 7 might be involved in the fourth code digit calculation , I`ve observed the following.

From the above four examples , the fourth code digit is a multiplication of 7 using the second Box Digit

eg from the above 3 x 7 = 21...so fourth code digit is 1

This only works with R0*2 box numbers ( ie Beginning with 0 and ending in 2 )

Presumably the 0 and/or the 2 of the Box Number and possibly the 8 from the Code Number are used in these examples but I can`t see a connection at the moment with the other Boxes/Codes in the main list.

Anyone have any ideas?.

[Guest]

Here is a code breaking problem set by my son`s Engineering Maths Lecturer as “something to keep your mind active” during the Summer Vacation…..and he can only solve half of it so far.

Can anyone help?

===========================================================

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 is several million of ££s .

Each box is labelled with a letter and three digits and the thief knows that the combination lock code for each individual box can be calculated from this sequence.

Eg ...........Box No T176 has a Combination Lock Code of 6896

However , each box has this 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.

The aim is to open two particular boxes which contain the most valuable bag of diamonds

So the question is .

"What is the code for boxes W456 and R588 ?"

Here is the list of open BOX numbers with their security CODES from inside.

Can you workout ( and explain ! ) the method to determine the combination lock code of Boxes W456 and R588

BOX CODE BOX CODE BOX CODE

P194 0851 P327 0606 P541 0427

Q068 9969 Q409 9511 Q638 9382

R002 8940 R012 8947 R017 8990

R025 8956 R030 8900 R031 8970

R032 8941 R033 8912 R037 8994

R054 8986 R082 8946 R111 8847

R125 8827 R142 8819 R150 8874

R163 8883 R216 8761 R228 8700

R283 8758 R291 8714 R294 8726

R342 8650 R370 8610 R395 8667

R437 8577 R544 8430 R560 8454

R568 8410 R642 8362 R679 8358

R711 8261 R754 8271 R808 8120

R836 8189 R846 8186 R854 8142

R870 8163 R876 8187 R941 8004

S526 7477 S702 7232 S982 7079

T337 6601 T668 6384 T860 6160

V120 4871 V555 4409 V950 4038

X482 2521 X613 2334 X623 2331

HELP !!

Anyone still looking? There is only Code Digit Four to work out now.

I am convinced that it is the result of a "multiply by 7 "....similar to Code Digit Three....but can not find the correct combination of other numbers to use.

I fancy Box Digit Two and Three with Code Digit One....but does anyone have any other theories?

Don`t leave me on my own with this one...p l e a s e ! !

It`s nearly CRACKED !

[Guest]

Well I got it.... but their has to be a better way of doing it. It took me like 5 hours and this is the best I could come up with.

To show that this works I'm going to give 4 examples

P327

R216

V555

X623

To start we use simple substitution for the first two numbers.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0

1 2 3 4 5 6 7 8 9 0

8 7 6 5 4 3 2 1 0 9

P327 = 06

R216 = 87

V555 = 44

X623 = 23

Next we add spot 2 and 4 of the unencrypted code and multiply them by 7

P327 = 06 , (3+7)*7=70

R216 = 87 , (2+6)*7=56

V555 = 44 , (5+5)*7=70

X623 = 23 , (6+3)*7=63

We place the last number into the encrypted code and keep the first for future use.

P327 = 060 , 7

R216 = 876 , 5

V555 = 440 , 7

X623 = 233 , 6

Next we add 5 to the first number of the encrypted code, and subtract the third number of the unencrypted code.

P327 = 060 , 7 , 0+5-2=03

R216 = 876 , 5 , 8+5-1=12

V555 = 440 , 7 , 4+5-5=04

X623 = 233 , 6 , 2+5-2=05

Discard the first number and multiply the second number by 7

P327 = 060 , 7 , 3*7=21

R216 = 876 , 5 , 2*7=14

V555 = 440 , 7 , 4*7=28

X623 = 233 , 6 , 5*7=35

Again discard the first number so we end up with this

P327 = 060 , 7 , 1

R216 = 876 , 5 , 4

V555 = 440 , 7 , 8

X623 = 233 , 6 , 5

Now subtract our two numbers add 10 and again discard the first number. The one left is the last number of the encrypted code.

P327 = 060 , 7-1+10=06

R216 = 876 , 5-4+10=11

V555 = 440 , 7-8+10=09

X623 = 233 , 6-5+10=11

P327 = 0606

R216 = 8761

V555 = 4409

X623 = 2331

So the answer to the riddle (if I did it right) is

W456

R588

W456 35 , (4+5)*7=63

R588 84 , (5+8)*7=91

W456 353 , 6 , 3+5-5=03

R588 841 , 9 , 8+5-8=05

W456 353 , 6 , 7*3=21

R588 841 , 9 , 7*5=35

W456 353 , 6 , 1

R588 841 , 9 , 5

W456 353 , 6-1=5

R588 841 , 9-5=4

W456 = 3535

R588 = 8414

I hope I did the math right on the final answer.

Edited August 16, 2008 by toxicoxyde

[Guest]

[Guest]

pmsl ..... Did not go thru it but it seems very complex

[Guest]

Hope redrooster is satisfied now and bravo to toxicoxyde for this effort

[Guest]

Well I got it.... but their has to be a better way of doing it. It took me like 5 hours and this is the best I could come up with.

To show that this works I'm going to give 4 examples

P327

R216

V555

X623

To start we use simple substitution for the first two numbers.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0

1 2 3 4 5 6 7 8 9 0

8 7 6 5 4 3 2 1 0 9

P327 = 06

R216 = 87

V555 = 44

X623 = 23

Next we add spot 2 and 4 of the unencrypted code and multiply them by 7

P327 = 06 , (3+7)*7=70

R216 = 87 , (2+6)*7=56

V555 = 44 , (5+5)*7=70

X623 = 23 , (6+3)*7=63

We place the last number into the encrypted code and keep the first for future use.

P327 = 060 , 7

R216 = 876 , 5

V555 = 440 , 7

X623 = 233 , 6

Next we add 5 to the first number of the encrypted code, and subtract the third number of the unencrypted code.

P327 = 060 , 7 , 0+5-2=03

R216 = 876 , 5 , 8+5-1=12

V555 = 440 , 7 , 4+5-5=04

X623 = 233 , 6 , 2+5-2=05

Discard the first number and multiply the second number by 7

P327 = 060 , 7 , 3*7=21

R216 = 876 , 5 , 2*7=14

V555 = 440 , 7 , 4*7=28

X623 = 233 , 6 , 5*7=35

Again discard the first number so we end up with this

P327 = 060 , 7 , 1

R216 = 876 , 5 , 4

V555 = 440 , 7 , 8

X623 = 233 , 6 , 5

Now subtract our two numbers add 10 and again discard the first number. The one left is the last number of the encrypted code.

P327 = 060 , 7-1+10=06

R216 = 876 , 5-4+10=11

V555 = 440 , 7-8+10=09

X623 = 233 , 6-5+10=11

P327 = 0606

R216 = 8761

V555 = 4409

X623 = 2331

So the answer to the riddle (if I did it right) is

W456

R588

W456 35 , (4+5)*7=63

R588 84 , (5+8)*7=91

W456 353 , 6 , 3+5-5=03

R588 841 , 9 , 8+5-8=05

W456 353 , 6 , 7*3=21

R588 841 , 9 , 7*5=35

W456 353 , 6 , 1

R588 841 , 9 , 5

W456 353 , 6-1=5

R588 841 , 9-5=4

W456 = 3535

R588 = 8414

I hope I did the math right on the final answer.

What an excellent result ! Thanks very much.

Yes it does work OK . Amongst the many, many different methods / systems attempted , I had tried a similar path by subtracting from 10....which only worked occasionally. I had never even considered using (ie ...completely overlooked !) the "tens" part of the previous Code Digit Three calculation which was the underlying variable to the final answer for this Code Digit Four

If you did it in only 5 hours , then "Well done"..We`ve spent hours+hours trying to solve this final digt.....and still didn`t find the correct method.

Just as a matter of interest, what procedure did you follow to lead you to this method as we`d be intrigued to know......for the next time !

[Guest]

I got the first 3 digits slightly differently, but still got there. Adding what I learned from Toxicoxyde I put it into a spread sheet. [i tried to upload, but the system won't let me. I'll send it if any one wants to look at it.]

2 things ....

the formulas work for all but 7 of the combinations "found" (it works so well for the rest, i figure there's a problem with those 7 pieces of data)

The spread sheet gets a different answer for W456.

Worked on this one a long time.... GOOD QUESTION!!!

[Guest]

What an excellent result ! Thanks very much.

Yes it does work OK . Amongst the many, many different methods / systems attempted , I had tried a similar path by subtracting from 10....which only worked occasionally. I had never even considered using (ie ...completely overlooked !) the "tens" part of the previous Code Digit Three calculation which was the underlying variable to the final answer for this Code Digit Four

If you did it in only 5 hours , then "Well done"..We`ve spent hours+hours trying to solve this final digt.....and still didn`t find the correct method.

Just as a matter of interest, what procedure did you follow to lead you to this method as we`d be intrigued to know......for the next time !

Well I first noticed it with this sequence

R030 8900 = 0+0*7 = 00

R031 8970 = 0+1*7 = 07

R032 8941 = 0+2*7 = 14

R033 8912 = 0+3*7 = 21

R037 8994 = 0+7*7 = 49

with the R03x series all you have to do is multiply box digits two and for together the swap the numbers to get the last two numbers of the code.

So I checked some other numbers and found

R437 8577 = (4+3)*7 = 77

R836 8189 = (8+3)*7 = 98

It worked the same with them. Now the only common parts where the R and the three. So I checked for other combination's and found this.

R163 8883 = (1+3)*7 = 28

R560 8454 = (5+0)*7 = 35

R568 8410 = (5+8)*7 = 91

I noticed that if you swap the digits and add 1 you get the last number of the code.

2+1 = 3

3+1 = 4

9+1 = 10

So I did the same thing with a few more numbers and created a truth table for the letter R.

R

Box digit 3:-------1-2-3-4-5-6-7-8-9-0

Number to add:--6-3-0-7-4-1-8-5-2-9

Note that the order is multiples of 7

(x7,x4,x1,x8,x5,x2,x9,x6,x3,x0)

Did the same thing to s and got this

S526 7477 = (5+6)*7 = 77 + 0

S702 7232 = (7+2)*7 = 63 + 6

S982 7079 = (9+2)*7 = 77 + 2

S

Box digit 3:-------1-2-3-4-5-6-7-8-9-0

Number to add:-----0-------------2---6

and noticed that it is the same as R just shifted over one.

Now the only thing left was to express this into a algorithm.

Code 1 = Substitution

Code 2 = Substitution

Code 3 = (Box2 + Box3)*7 (Then swap)

Code 4 = (Remaining number)+((Code1+5-Box3SecDigit)*7)

(Note: the only reason I have a +10 in my original solution is to keep the number form going negative)

And that's how I came to my answer.

Edited August 18, 2008 by toxicoxyde

[Guest]

I got the first 3 digits slightly differently, but still got there. Adding what I learned from Toxicoxyde I put it into a spread sheet. [i tried to upload, but the system won't let me. I'll send it if any one wants to look at it.]

2 things ....

the formulas work for all but 7 of the combinations "found" (it works so well for the rest, i figure there's a problem with those 7 pieces of data)

What 7 didn't it work for? My answer could be wrong... I could be missing something.

The spread sheet gets a different answer for W456.

What answer did you get?

I would also be interested to know what you came up with for the first three parts of the code. I always like to hear about different approaches.

[Guest]

I would also be interested to know what you came up with for the first three parts of the code. I always like to hear about different approaches.

I realised a connection with first box digit and third box digit always added together to the same resulting last digit ...but this was not the third code digit.

Writing it all out , I eventually realised that the relationship depended on a multiplication of 7 which , using the last digit in this answer ,always gave the third code digit .

[Guest]

Update: There have been two more different suggestions in the method used to calculate Code Digit Four.

See the second part of this problem

Title :

New Code Breaker Wanted (Part TWO)