While working in your basement, you notice there is a portion of concrete which is not like the rest. It is funny you never noticed it before, but most of the time there was a rug covering it. Tapping on the area made a somewhat hollow sound, so hammer in hand, you smash into the concrete... Hands ringing, but hole made you see an object hidden under the concrete.

Getting a bigger hammer, you open up the hole more to allow for the object to be removed, and then pull out a small, but heavy trunk. There is a lock on it with 6 digit places and what appears to be the numbers 0-9 inclusive.

Inspecting the trunk, there is writing all over it. You notice 6 distinct strings of characters, but there is no indication of the order of the strings. These must have something to do with the contents, the combo, or both!

  • zoxluycqpexgggbdthnwoxqkp
  • ujzhpvklmrurndrdkmhrasklm
  • ykswutxrjjluvrqvkljwdrhxx
  • njjkjdwcciuentkstgcrthkws
  • dmbequmdzgytgifbzgntphjtd
  • mnwwhkjuqjijdoibrikevxntc

Hint 1

The characters within each string can be re-arranged and arrive at the same result...

After puzzling over this trunk and strings, you look at the lock a bit closer. Hardly noticeable under the grime is the inscription $27$, and on the other side is inscribed $10$. Once you clean up the lock as best you can, you also notice a faint inscription on the lock bar... Count the strings, you'll do just fine... make your math no greater than 9. Strange indeed...

What is the combination, and what is inside? (2 answers needed)

  • 1
    $\begingroup$ Thank you for the tag update. I knew there had to be another one. $\endgroup$ May 25, 2017 at 21:42
  • $\begingroup$ What do you mean by "piece together"? Were the characters grouped together in some way, or are the groups completely random? $\endgroup$
    – MikeQ
    May 26, 2017 at 19:47
  • $\begingroup$ Updated puzzle to reflect the answer, and included a hint. $\endgroup$ May 26, 2017 at 20:00
  • $\begingroup$ I would appreciate input as well as to the quality of this puzzle and what would make it better... $\endgroup$ Jun 9, 2017 at 19:27

4 Answers 4


My guess is:


The clue "Count the strings" seems to indicate that the values of each character in each string should be summed.

Using the ASCII values of each character, the strings have values {2763, 2742, 2827, 2716, 2706, 2725}, in their original order (this may satisfy the inscribed "$27$"). Using the alphabetical position of each character (i.e. a = 1, b = 2,...,z = 26), we have {363, 342, 427, 316, 306, 325}, respectively. Note that adding 2400 to each element in the second array yields the first array. Since there are 6 strings, it's reasonable to assume that each string corresponds to a single digit in the combination.

Now, the inscription indicates "make your math no greater than 9," which seems to suggest that...

each value should be considered (mod 10), which would satisfy the inscribed $10$ and yield 6 digits between 0 and 9. Regardless of whether the ASCII values or alphabetical positions are used, the last digits of each string value are {3, 2, 7, 6, 6, 5}.

The problem mentions that "no indication of the order of the strings" which may suggest that we must introduce order to the strings ourselves.

Thus, sorting the string values yields {2706, 2716, 2725, 2742, 2763, 2827} which translates to {6, 6, 5, 2, 3, 7}, or 665237. Another possible answer might be 235667, if sorting is performed after taking each value mod 10.

  • $\begingroup$ You are on the right path... keep it up! Your interpretation of 27 is not quite right, but you are getting there. I updated the last string, so do the math again on it. +1 for effort so far! Also, remember there are two answers I am looking for. $\endgroup$ Jun 9, 2017 at 16:05
  • $\begingroup$ Instead of mod-ing it, try recursive adding of numerals in the resulting set until it's a single digit. $\endgroup$ Jun 15, 2017 at 14:35

As I am quitting Puzzling.SE, here is the answer.

First step:

Convert each string from letters to numbers in the pattern a=1 as was said above.


add them up as was done above by Joel Abraham. The totals are {363, 342, 427, 316, 306, 325}


Fulfill the 27 by mod27 on the sums from above. You get {12, 18, 22, 19, 9, 5}


Convert back to letters and then find the anagram. {L R V S I E} = {SILVER} = contents, which also points to the right order of the next step.


Take the previous sums and apply mod10 (make no larger than 9) to get the actual combo, just out of order. {3, 2, 7, 6, 6, 5} Re-arrange according to the order SILVER making {6, 6, 3, 7, 5, 2} which is the combination.

Final Answer is:

The combination is 663752 and the contents are SILVER


I might be crazy but a simple guess at the combo is



count the strings ->


is 6 (first digit)


make your math no greater than 9 (last digit)


numbers 6-9 (6789) could be 4 digit combo.

Curious if conversion from base10 to baseN could be involved.

  • $\begingroup$ Don't give up! Remember, it is a 6 digit lock, so you will need more digits, and regarding conversion, there is one conversion which all the rest depend upon... $\endgroup$ Jun 9, 2017 at 19:28

I think someone else has already gotten closer, but I decided to go ahead and count up the instances of each letter in case it does any good.

# of times each letter occurs

1: af
3: oy
4: bepsvz
5: cilq
6: m
7: ghnwx
8: u
9: drt
11: jk
  • $\begingroup$ I would pay attention to the thinking of Joel Abraham... he is on the right track. Don't give up! $\endgroup$ Jun 21, 2017 at 18:56

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