Below is a puzzle called crack the lock. The rules are simple:

  • the lock contains 16 characters in this particular puzzle
  • No character can be used twice
  • the two numbers on the the far right tell you two things
  • the first number tells you how many of the characters in each line are used in the final answer
    • Ex: for the first line 10 of the characters out of the 16 in that line will be used in the answer.
  • The second number tells you how many of the characters used are in the correct place
    • Ex: again for the first line we know 10 of the characters will be used and of those 10, 8 of them are in the correct spot.

K B C 8 2 W % T + 7 E R 4 9 6 A 10 C, 8 R
A K T 1 % 3 R @ W D M F 8 5 9 7 9 C, 8 R
K B F 8 2 W 4 T + 7 5 R C 9 1 A 10 C, 3 R

(As an image)

The logic used here is somewhat different than the other "crack the lock questions" I've solved. How do I start this puzzle?

If interested in more puzzles like this or this puzzle specifically look for @ali_imashli on TikTok where he regularly posts puzzles like this. This is the link to the direct post, second picture is this question but be warned the answer may be in the comments

  • $\begingroup$ (Some comments here deleted) @Rcyr1010 please remember that comments are for clarifying questions, exactly what had happened here. Many puzzles hinge on correctly understanding details, and the difference between a number and a digit is not “minor semantics”, it’s an important distinction for understanding the puzzle. Graciously accepting corrections and suggestions in the spirit in which they were given helps make better and more satisfying puzzle presentations. $\endgroup$
    – Rubio
    Commented Aug 17, 2023 at 9:20

4 Answers 4


The main thing you need to know to solve this puzzle is that

the first two codes have a total of 16 symbols in the correct places.
However, while the two codes have several symbols in common, none are in the same position. Therefore, the answer consists of 8 positions of the first code and the other 8 positions of the second code.

[Solution path below]

the third code differs from the first in exactly five places. Since it also has five fewer correct positions, those five positions in the first code must be correct. However, the corresponding symbols in the third code must still be in the answer, since it has the same number of correct symbols, so those are placed in their positions from the second code.
Progress so far: --C1--%--EF456-

Notice now that

the second code only has one correct symbol in the wrong position, and it's %. Therefore, for any other position in the first code where the same symbol is elsewhere in the second code, the second code has that position correct.
We have now placed the second code's eight correct positions, so the remainder of the answer can be pulled from the first code: ABC123%@+DEF4567

  • $\begingroup$ Thank you for your help! $\endgroup$
    – Rcyr1010
    Commented Jul 31, 2023 at 4:19
  • $\begingroup$ So I worked through this using your method and I understand how to get through the first two steps but the last step doesn't make sense to me. 8 can't understand why k, t, r, w, 8, 9 wouldn't be in the answer then according to your third step can you elaborate a bit? $\endgroup$
    – Rcyr1010
    Commented Aug 2, 2023 at 19:39
  • 1
    $\begingroup$ Consider the fourth spot: Either the first or second code has that spot correct. If it's the first code, then the second code has at least two correct symbols in the wrong places: 8 and %. Since the second code only has one, this can't happen, and the second code is the correct one. $\endgroup$ Commented Aug 4, 2023 at 17:00

1. Look for the most obvious commonalities first:

enter image description here

Since $8-3=5$ and that covers all the remaining characters, all of C, %, E, 4 and 6 must be in their correct place. Conversely, F, 4, 5, C and 1 are correct, but in the wrong place because the first and third guesses have the same number of correct characters.

2. enter image description here

On the third line, % is correct but placed wrong. All the remaining correct characters must also be placed correctly there, so T, R, M, 8 and 9 can't be in the code at all. For the same reason, 1, F and 5 are correctly placed on the last line.

3. enter image description here

The first line's correctly placed characters can only be B, 2 and +.

enter image description here

Something of a return to 2. We can figure out that K and W aren't in the code, so only 7 and A are left on the first two lines.

5. Conclusion:
enter image description here

The code is ABC123%@+DEF4567.


I solved it in a rather different way, not using the 10c info from the third line.

First as AxiomaticSystem noted:

The first two clues share no characters in the same positions and therefore there's no overlap in their correct positions. Since their total correct positions totals to 16, that covers all of the answer.

Next, I noted that the first line only has two characters that are in the answer, but in the wrong position, and the second line only has one character that's in the answer that's in the wrong position. That means any characters that are shared between the two lines are either not in the answer, or must be counted as one of these three characters. Then I marked all of these characters.

common characters in lines 1 and two circled in red

Since there are three positions where both lines have one of these characters in common, one must be picked for each of those positions. That means that none of the other circled ones can be in the correct position. Thus the character from the other line must be in the correct position:

correct positions circled

From those correct characters we can mark three characters from the bottom line correct as well. Since there are only three characters in the bottom line that are in the correct position, we know that the leading K and trailing A must be wrong.

Now the only there's only 2 positions unsolved, and the top line still needs two more correct, so they must be from the top line:

final answer

  • $\begingroup$ Clever solution! $\endgroup$ Commented Jul 18, 2023 at 10:59
  • $\begingroup$ Thank you for this idea! I am trying to solve more puzzles like this. I appreciate the idess $\endgroup$
    – Rcyr1010
    Commented Jul 31, 2023 at 4:20

The asker seems to be looking more for a method than for an answer. My method was a bit different to everyone else's, so even though it comes out to the same answer I felt it would be appropriate to write it up. Hopefully someone finds it useful.


For this answer, "Rule" will mean a set of up to 16 characters in specific locations, combined with two numbers representing the Correct and Rightly-placed characters in that set. We begin this puzzle with three Rules.

Rule 1: K B C 8 2 W % T + 7 E R 4 9 6 A         10/8
Rule 2: A K T 1 % 3 R @ W D M F 8 5 9 7         9/8
Rule 3: K B F 8 2 W 4 T + 7 5 R C 9 1 A         10/3

The method I used to solve this puzzle was to compare and combine the Rules we have in order to create new Rules. I am aiming to create perfect Rules: rules that contain only correct and rightly placed characters. The closer a rule is to being perfect, the more useful it is. For example:

Recognising that Rules 1 and 3 are similar, we can separate the commonalities from the differences to create 3 new rules from those two.
Rule 4: K B 8 2 W T + 7 R 9 A from both
Rule 5: C % E 4 6 from R1
Rule 6: F 4 5 C 1 from R3
Since no more than 3 characters in Rule 4 can be rightly-placed (from Rule 3), all 5 characters in Rule 5 must be rightly-placed (from Rule 1),
Rule 4: K B 8 2 W T + 7 R 9 A 5/3
Rule 5: C % E 4 6 5/5
and no characters in Rule 6 can be rightly-placed, though they must all be correct
Rule 6: F 4 5 C 1 5/0

Note that Rule 5 is a perfect rule: it has 5 characters in the set, and all 5 of them are correct and rightly-placed. To solve this puzzle, just continue experimenting with breaking down the rules you have into rules that are closer to perfect until you get to the answer.

Answer, continued

From here, we can also note that from Rules 5 and 6 we know 8 of the 16 correct characters. Separating the characters we know must be in the final result from Rule 2, we can create two new Rules:

Rule 7: A K T 3 R @ W D M 8 9 7 5/5
Rule 8: 1 F 5 3/3
Rule 8.5: % 1/0
5 characters in Rule 7 must be Correct, since we have removed 4 correct characters from Rule 2. No more than 3 characters in Rule 8 can be Rightly-placed, so 5 or more characters in Rule 7 must be Rightly-placed to match '8R' from Rule 2. Therefore, logically, Rule 7 must have 5 Correct and Rightly-placed characters, and Rule 8 becomes perfect with 3 Correct and Rightly Placed characters.

Now, using both Rule 5 and Rule 8, we have 8 characters correct and rightly-placed. We can get the rest by comparing and combining Rules 4 and 7 to create 3 new rules:

Note that Rules 4 and 7 share no characters with the 8 that we know from Rules 5 and 8. They contain only the 8 unknown characters, and must contain all of them. Between them they have 10 correct characters, so two of them must be shared between the two rules. Rule 4 has 3 rightly-placed characters, and Rule 7 has 5, so the shared characters will be rightly-placed in Rule 7.
Separate the shared characters into one rule with placement from Rule 7, and the not-shared characters into two other rules:
Rule 9 : A K T W 8 9 7 2/2 from shared chars
Rule 10: B 2 + R 3/3 from R4
Rule 11: 3 R @ D M 3/3 from R7

We now have 5 rules which contain all 16 characters correct and rightly-placed. Compare them and resolve the overlaps:

Rule 5 : C % E 4 6 5/5
Rule 8 : 1 F 5 3/3
Rule 9 : A K T W 8 9 7 2/2
Rule 10: B 2 + R 3/3
Rule 11: 3 R @ D M 3/3
Rules 5, 8, 10 and 11 resolve simply. Rules 5 and 8 are perfect (containing no incorrect characters) and Rules 10 and 11 become perfect after removing the characters that overlap R5/8: (R, R and M)
Rule 12: B C 1 2 3 % @ + D E F 4 5 6 14/14
Rule 9 : A K T W 8 9 7 2/2
Rule 9 can be perfected by removing overlaps with Rule 12, and the puzzle is solved in Rule 13:
Rule 13: A B C 1 2 3 % @ + D E F 4 5 6 7 16/16

  • $\begingroup$ This is very interesting, thank you!! $\endgroup$
    – Rcyr1010
    Commented Jul 31, 2023 at 4:20

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