[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?

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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 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]

p=0

q=9

r=8

s=7

t=6

u=5

v=4

w=3

x=2

y=1

working on the numbers now.. they look a little harder

[Guest]

nope i think it is wrong, silly me

Edited August 7, 2008 by yellowsubmarine

[Guest]

the 1st number after the letter is

0=9

1=8

2=7

3=6

4=5

5=4

6=3

7=2

8=1

9=0

[Prof. Templeton]

working on the numbers now.. they look a little harder

p=0

q=9

r=8

s=7

t=6

u=5

v=4

w=3

x=2

y=1

There are a lot of "R" boxes and it seems that that numbers corrulate the same for all the "R"'s eg.

0=9

1=8

2=7

3=6

4=5

5=4

6=3

7=2

8=1

9=0

But I think only in the second position.

Just brainstorming.

[Guest]

0=9

1=8

2=7

3=6

4=5

5=4

6=3

7=2

8=1

9=0

[Guest]

1st row

0-9

1-8

2-7

3-6

4-5

5-4

6-3

7-2

8-1

9-0

[Guest]

stumped at the min with the last 2 numbers though

[Guest]

I noticed the following:

R0x2 will always yield 89 4x

So if you look at all the samples that start with R0:

2 corresponds to 4

0 - 0

1 - 7

2 - 4

3 - 1

4 - 8

5 - 5

6 -

7 - 9

8 - 6 *

9 -

* This result is from Q

At the beginning, I thought the number squared and only the last digit, but doesn't fit everything....

Edited August 7, 2008 by v333k

[Guest]

I think I have the 3rd column

Let me know if you think its wrong, or if you agree

Take the 2st and 4th box numbers and add them and look at the 1s column.

If something is > 9 (IE 4+9=13 you use the 3 the single digits)

Sum(2nd +4th)=number

0=0

1=7

2=4

3=1

4=8

5=5

6=2

7=9

8=6

9=3

[Guest]

I think I have the 3rd column

Let me know if you think its wrong, or if you agree

Take the 2st and 4th box numbers and add them and look at the 1s column.

If something is > 9 (IE 4+9=13 you use the 3 the single digits)

Sum(2nd +4th)=number

0=0

1=7

2=4

3=1

4=8

5=5

6=2

7=9

8=6

9=3

example?

[Guest]

I think I have the 3rd column

Let me know if you think its wrong, or if you agree

Take the 2st and 4th box numbers and add them and look at the 1s column.

If something is > 9 (IE 4+9=13 you use the 3 the single digits)

Sum(2nd +4th)=number

0=0

1=7

2=4

3=1

4=8

5=5

6=2

7=9

8=6

9=3

Yep, I tested it. It works on all the open boxes. nice work!

Q409: 4+9=13, use the 3, 3=1, so the 3rd number is 1

Edited August 7, 2008 by Cherry Lane

[Guest]

I think I have the 3rd column

Let me know if you think its wrong, or if you agree

Take the 2st and 4th box numbers and add them and look at the 1s column.

If something is > 9 (IE 4+9=13 you use the 3 the single digits)

Sum(2nd +4th)=number

0=0

1=7

2=4

3=1

4=8

5=5

6=2

7=9

8=6

9=3

Looks convincing?

[Guest]

I think I have the 3rd column

Let me know if you think its wrong, or if you agree

Take the 2st and 4th box numbers and add them and look at the 1s column.

If something is > 9 (IE 4+9=13 you use the 3 the single digits)

Sum(2nd +4th)=number

0=0

1=7

2=4

3=1

4=8

5=5

6=2

7=9

8=6

9=3

yes it is right - the correlation ios found by multiplying with 7 and keeping the units digit.. So 7x1 = 7, 7x2=14 (4) and so on..

[Guest]

Yep, I tested it. It works on all the open boxes. nice work!

Q409: 4+9=13, use the 3, 3=1, so the 3rd number is 1

So all that`s left is the fourth digit.

Nearly there !

[Guest]

for R588 i have:

84?0

[Guest]

for R588 i have:

84?0

The 3rd digit system has already been deduced....so no need for ? in your query

[Guest]

R588=8460

[Guest]

R588=8460

R588..Using the solution to the 3rd digit .( 5 + 8 =13..using the 3x7 = 21 ..so 3rd digit is 1 )

Your solution then becomes 8410.

Can you explain the theory behind your 4th digit ?

[Guest]

to be honest i got that before all the stuff on 8+13=3+1 etc and i saw a R box with 8 as the last letter and assumed that that was it, it may not be correct i hope it is.

don't blame me, im 13!

[Guest]

to be honest i got that before all the stuff on 8+13=3+1 etc and i saw a R box with 8 as the last letter and assumed that that was it, it may not be correct i hope it is.

don't blame me, im 13!

I don`t think it is

[Guest]

Only one more set of figures to find!

Has everyone given up on this one?

[Guest]

Only one more set of figures to find!

Has everyone given up on this one?

I just hope you are not doing your son's work for him.

>>From my years of teaching.

[Guest]

Just a recap of what we know,

we'll use the 1st example...

P194 0851

OK so...

P=0<~~~

Q=9

R=8

S=7

T=6

U=5

V=4

W=3

X=2...

1-9=8<~~~ subtract 9 from the 1st digit to get your 2nd column

1+4=5 and that = (5) since...

[0=0

1=7

2=4

3=1

4=8

5=5<~~~

6=2

7=9

8=6

9=3

anything greater than 9 you only use the single digit 9+9=18 so the 8 is the digit we use which means the number is 6]

Here is where i'm stuck...was hoping these consecutive codes could help but nothing so far...

R030 8900

R031 8970

R032 8941

R033 8912

...

R037 8994

[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?

[Guest]

I just hope you are not doing your son's work for him.

>>From my years of teaching.

Not really helping him with any "homework" as he is doing further post graduate studies at University and this is supposed to be some form of light entertainment/ holiday readingto keep their mental abilities alert!

This is only one exercise amongst many others.