The London safes and their secret combinations

Question

Here's the first riddle I invented for Puzzling SE. Hope you'll enjoy solving it.

This story takes place during World War II. A German spy is assigned the task to sneak overnight into a military base near London and steal secret documents of utmost importance.

The documents are kept inside ninety safes of reinforced unbreakable steel, and each safe is protected by a numeric secret code. The only way to get the documents is to open each safe with the correct code. If the wrong code is entered, the safe will immediately trigger a siren which will alert all military personnel, so the spy has only one try available.

German Intelligence was able to intercept the codes of the first fifteen safes, which they communicate to the spy:

Safe   Code
#1     2
#2     4
#3     6 
#4     8
#5     10
#6     12
#7     14
#8     16
#9     18
#10    20
#11    22
#12    24
#13    26
#14    28
#15    30

The rest of the safes' codes is still unknown. Therefore German Intelligence advises the spy not to tamper with the other safes.

That same night, the spy enters unseen the military base, finds the safes, silently opens each one of the first fifteen safes with the correct code and steals the documents.

However, wanting to steal as many secret documents as possible, the spy starts opening the rest of the safes. The pattern is obvious: a safe's code appears to be the double of the safe's number!

Everything goes well for safe #16 and #17.

However, when the spy enters the code 36 on safe #18, the safe doesn't open; its alarm triggers, and the spy is captured.

Can you find why? If not, continue reading.

After the end of the war, the commander of the base is asked by the High Command to compile a list of all safes and codes. He produces the following document:

Safe   Code   Safe   Code   Safe   Code   Safe   Code   Safe   Code   Safe   Code
#1     2      #16    32     #31    28     #46   -42     #61    34     #76    46     
#2     4      #17    34     #32    30     #47   -40     #62    36     #77    48    
#3     6      #18    25     #33    32     #48   -38     #63    38     #78    50
#4     8      #19    38     #34    34     #49   -36     #64    40     #79    52   
#5     10     #20    40     #35    36     #50    30     #65    42     #80    25    
#6     12     #21    42     #36    38     #51    32     #66    44     #81    27    
#7     14     #22    44     #37    40     #52    34     #67    46     #82    29
#8     16     #23    46     #38    42     #53    36     #68    48     #83    31
#9     18     #24    48     #39    44     #54    38     #69    50     #84    33
#10    20     #25    50     #40   -54     #55    40     #70    34     #85    35
#11    22     #26    52     #41   -52     #56    42     #71    36     #86    37
#12    24     #27    54     #42   -50     #57    44     #72    38     #87    39
#13    26     #28    56     #43   -48     #58    46     #73    40     #88    41
#14    28     #29    58     #44   -46     #59    48     #74    42     #89    43
#15    30     #30    26     #45   -44     #60    32     #75    44     #90    38

Can you find now how the codes were chosen - and hence why the spy failed?

Hint 1

There is an unique solution for the riddle. Don't try to find overcomplicated formulas. There is more than just mathematics -- in fact, although there is quite a bit of calculations, the mathematics involved are very basic. To crack the code, you'll have to do some lateral thinking.

Hint 2

Language is important. The story takes place in Great Britain because the riddle is related to the English language. The nationality of the spy is irrelevant.

Hint 3

Observe how the British chose the secret algorithm to fool spies. "Code of safe $n = 2 \times n$" is a red herring -- it looks like it is the solution, but is valid only up to safe #29 with the exception of safe #18. The units for each "ten" follow a pattern, but each "ten" breaks the pattern. Where would you find such a pattern?

EDIT: simplified the puzzle story.

Answer 1

To get the code of a safe, write down the safe's number as a word, replace the letters with numbers using the following matching, and finally add them together.

 e: 0    f: 5    g: -62  h: 67
 i: 0    l: 8    n: 9    o: -7
 r: -72  s: 0    t: 11   u: 82
 v: 5    w: 0    x: 12   y: 9 

(the other letters do not appear in the numbers)

How I got there:

After the last hint (and all the comments and thinking before) it was pretty clear to me, that each letter in the word had to have some value which then would add up to the safe's code. Obviously the usual matching a = 1, b = 2, ... would not fit, so I had to do some calculations based on the assumption that t+y = 20 which I guessed after hint 5.

 t+y = 20
 t+w+e+n+t+y = 40 => t+w+e+n = 20
 t+e+n = 20 = t+w+e+n => w = 0
 t+w+o = t+o = 4, t+e+n = 20 => t+t+o+n+e = 24
 t+t+o+n+e = 24, o+n+e = 2 => t+t = 22 => t = 11
 t+y = 20, t = 11 => y = 9
 t+e+n = 20, t = 11 => e+n = 9
 e+n = 9, o+n+e = 2 => o = -7
 n+i+n+e = 18, e+n = 9 => n+i = 9 => i = e
 n+i+n+e+t+e+e+n = 38, n+i+n+e = 18 => t+e+e+n = 20
 t+e+e+n = 20 = t+e+n => e = i = 0, n = 9
 f+i+f+t+y = 30, t+y = 20 => f+i+f = f+f = 10 => f = 5
 f+i+v+e = f+v = 10 => v = 5
 s+e+v+e+n = s+v+n = 14 => s = 0
 e+l+e+v+e+n = l+v+n = 22 => l = 8
 s+i+x = s+x = 12 => x = 12
 f+o+r+t+y = -54 => r = -72
 f+o+u+r = 8 => u = 82
 t+h+r+e+e = t+h+r = 6 => h = 67
 e+i+g+h+t = g+h+t = 16 => g = -62
 

I hope that helps to understand the way I found the solution and also how the codes are calculated.

Answer 2

The code should have been

0

I believe that the codes for each safe are equivalent to

$($The next safe # in the sequence $\times 2) -2$. Since Safe 30 is at the end of the sequence, technically Safe 1 would be next in sequence $(1 \times 2) -2 = 0$

Answer 3

The code should have been

2

Explanation:

The story plays in the leap year 1944. The British picked the codes day by day, and simply doubled the number of the current day. The numbering started on Feburary 1 (code $2*1=2$), Feburary 2 (code $2*2=4$), $\ldots$, Feburary 29 (code $2*29=58$), March 1 (code $2*1=2$).

Answer 4

Partial Answer

The code has something to do with the digits of the safe number. If the code to safe $n$ is denoted by $C(n)$, then we can see that:

$$ C(n) = 2\left[f\left(\left\lfloor\frac{n}{10}\right\rfloor\right) + (n~\text{mod}~10)\right] $$

(The multiplication by two is because all code values are even.)

We know the values of $f$ for $0$ through $6$:

$$ \begin{align} f(0)&=0\\ f(1)&=10\\ f(2)&=20\\ f(3)&=13\\ f(4)&=-27\\ f(5)&=15\\ f(6)&=16\\ f(7)&=17 \end{align} $$

We are hinted that the answer has to do with the location of the safes, which leaves me to believe that $f$ is related to the English spelling or pronunciation of the numbers. In particular, it seems to me as if there is some relationship between thirty/thirteen, fifty/fifteen, etc., although $f(4)$ breaks this pattern.

Answer 5

I think this is too broad.

Anyway, the code might have been

0 or 1.

According to the formula:

$code=2*num\pmod{59}$
or
$code=2*num\pmod{60}$

Answer 6

I think I know how the code works but can't figure out the critical bits. Basically if you assume that values one to twenty are C(n)=2*n then every safe after that has a code that is equal to the sum of the codes of the constituent parts i.e. code(twentyone) = code(twenty)+code(one) code(thirtyone)=code(thirty)+code(one) and so on It would be nice to know how code(thirty), code(forty), code(fifty) and code(sixty) are calculated

Answer 7

I think the code is

62

Based on the idea that the code is

Each safe number ($n$) is mapped onto double the $n$th number that is not the sum of 4 distinct nonzero squares. This would mean the 30th safe was mapped to 62.

It could also be that the code is

Each safe number ($n$) is mapped onto double the $n$th number that has 2 or fewer distinct prime factors. Then $n=1,\ldots,29$ maps to $2n$ and 30 maps to 62.

Or it could be:

Each safe numbered ($n$) maps to double the $n$th number that cannot be expressed in form $p+q^2+r^3+s^4$, where $p, q, r, s$ are primes. Then $n$ maps to $2n$ for $n\in{1,\ldots,29}$ and 30 maps to 64

Obviously I used oeis.org to find the above, but it does suggest that the question is a bit broad

Answer 8

An answer without pure math

Safe number 1 starts with code = 2. Stepsize between safes is 2, except when exception fits as described hereunder.

Exeption-check:

When safe number written as English word ends with "y", replace "y" by "een" and look for that English word in the previous safe numbers. If found, take the code of that safe number. If not found, but there exists a word with a difference of only one letter, the following rule will apply:

• fetch previous replacement's safe number, i.e. of the "een" word (not safe's code)
• substract this safe number from actual safe number
• take code of the safe found this way
• multiply by -1 (because initial search was negative)

Answer 9

The code should have been

26

Partial explanation

Each ten breaks the pattern, except 1-29. Instead of the expected thirty * 2 the code uses thirteen (very similar in pronunciation): thirteen * 2 = 26. The same applies for fifty fifteen * 2 = 30, sixty sixteen * 2 = 32, seventy seventeen * 2 = 34, ninety nineteen * 2 = 38. This didn't work for eleven and twelve, which is why safes #1-29 follow a regular pattern.

However, I'm still in the dark on #40 and #80.