crack this code #1965












14












$begingroup$


Using the clues provided, determine the 'password'.



501 — Two correct numbers in wrong places
135 — One correct number in the right place
483 — All numbers are wrong
167 — One correct number in wrong place
430 — One correct number in the wrong place










share|improve this question











$endgroup$








  • 1




    $begingroup$
    +1 for reminding me that I haven't played Mastermind in a while...such a fun game.
    $endgroup$
    – Brandon_J
    Jan 29 at 19:06






  • 1




    $begingroup$
    While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :)
    $endgroup$
    – Ian MacDonald
    Jan 29 at 21:10
















14












$begingroup$


Using the clues provided, determine the 'password'.



501 — Two correct numbers in wrong places
135 — One correct number in the right place
483 — All numbers are wrong
167 — One correct number in wrong place
430 — One correct number in the wrong place










share|improve this question











$endgroup$








  • 1




    $begingroup$
    +1 for reminding me that I haven't played Mastermind in a while...such a fun game.
    $endgroup$
    – Brandon_J
    Jan 29 at 19:06






  • 1




    $begingroup$
    While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :)
    $endgroup$
    – Ian MacDonald
    Jan 29 at 21:10














14












14








14


1



$begingroup$


Using the clues provided, determine the 'password'.



501 — Two correct numbers in wrong places
135 — One correct number in the right place
483 — All numbers are wrong
167 — One correct number in wrong place
430 — One correct number in the wrong place










share|improve this question











$endgroup$




Using the clues provided, determine the 'password'.



501 — Two correct numbers in wrong places
135 — One correct number in the right place
483 — All numbers are wrong
167 — One correct number in wrong place
430 — One correct number in the wrong place







mastermind






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Feb 1 at 9:46









JonMark Perry

20k64096




20k64096










asked Jan 29 at 15:42









Brenda ScottBrenda Scott

713




713








  • 1




    $begingroup$
    +1 for reminding me that I haven't played Mastermind in a while...such a fun game.
    $endgroup$
    – Brandon_J
    Jan 29 at 19:06






  • 1




    $begingroup$
    While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :)
    $endgroup$
    – Ian MacDonald
    Jan 29 at 21:10














  • 1




    $begingroup$
    +1 for reminding me that I haven't played Mastermind in a while...such a fun game.
    $endgroup$
    – Brandon_J
    Jan 29 at 19:06






  • 1




    $begingroup$
    While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :)
    $endgroup$
    – Ian MacDonald
    Jan 29 at 21:10








1




1




$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 at 19:06




$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 at 19:06




1




1




$begingroup$
While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :)
$endgroup$
– Ian MacDonald
Jan 29 at 21:10




$begingroup$
While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :)
$endgroup$
– Ian MacDonald
Jan 29 at 21:10










9 Answers
9






active

oldest

votes


















11












$begingroup$

[5, 0, 1]: 2  
[1, 3, 5]: 1



So either 5 or 1 is correct, but not both.




[4, 8, 3]: none.

[1, 6, 7]: 1



Because 1 is in the same place as it was before, but the "one correct" is in the wrong place, it cannot be a right number. This means that 5 (from earlier) is definitely the last number, and also one of either 6 or 7 is correct, but in the wrong place.




[4, 3, 0]:



We know from above that 4 and 3 are both wrong, and we deduced from the last step that 1 was not a number, meaning the first line shows us that 5 and 0 are both numbers, so this actually gives us no new information on which numbers, but does tell us that 0 is still in the wrong place.



This means that we have "0 x 5" with x being either 6 or 7. From the second-last line, we know that the correct number is in the wrong place, so it can't be 6 (which was in the middle).



The final answer is: 0 7 5.







share|improve this answer











$endgroup$













  • $begingroup$
    I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)
    $endgroup$
    – Kevin
    Jan 29 at 22:20



















3












$begingroup$

The following are my musings to find the answer:



Starting with 430 (Rule 5):




4?? Eliminated by rule 5
?4? Eliminated by rule 3
??4 Eliminated by rule 3
3?? Eliminated by rule 3
?3? Eliminated by rule 5
??3 Eliminated by rule 3
0?? Remaining - Now Rule 6
?0? Remaining - Now Rule 6
??0 Eliminated by rule 5




Starting with 501 (Rule 1):




5?1 Eliminated by rule 1
51? Eliminated by rule 1
15? Eliminated by rule 6
1?5 Eliminated by rule 2
?15 Eliminated by rule 6
?51 Eliminated by rule 1
50? Eliminated by rule 1
5?0 Eliminated by rule 1
?50 Eliminated by rule 2
?05 Eliminated by rule 1
0?5 Remaining - Now Rule 7
05? Eliminated by rule 2
?01 Eliminated by rule 1
?10 Eliminated by rule 2
1?0 Eliminated by rule 6
10? Eliminated by rule 1
01? Eliminated by rule 2
0?1 Eliminated by rule 1




Starting with 167 (Rule 4):




1?? Elimated by rule 4
?1? Elimated by rule 1
??1 Elimated by rule 7
6?? Elimated by rule 7
?6? Elimated by rule 4
??6 Elimated by rule 7
7?? Elimated by rule 7
?7? Remaining - Now Rule 8
??7 Elimated by rule 4




Combining the two new rules, 7 and 8 gives:




Result 075







share|improve this answer









$endgroup$





















    2












    $begingroup$

    I've found the answers too complicate, so I share mine. Only three rules are needed:



    501 — Two correct numbers in wrong places




    So, only one of those numbers should be removed. Which one?




    135 — One correct number in the right place




    Since this shares 2 numbers with the first one, the 3 cannot be the
    correct one. (If the 3 comes into play, the rule would give info about
    2 numbers instead of one). So, from the previous rule, we should
    remove only the 1 or the 5. Now we know the number has a 0 in one side,
    and starts with 1 or end with 5. Only 2 chances: 1_0 or 0_5.




    167 — One correct number in wrong place




    This number start with 1 as the previous one, and the 1 can't be in
    right and wrong place at the same time. So, we have only one
    possibility now: 0_5. We are only missing the middle number and can't be the
    6 because it's already in the middle and not in the right place. So
    the only option left is the 7. -> 075.







    share|improve this answer









    $endgroup$













    • $begingroup$
      Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
      $endgroup$
      – North
      Jan 30 at 3:06










    • $begingroup$
      Excellent observation about the 3 to get it down to three rules!
      $endgroup$
      – par
      Jan 30 at 20:31





















    2












    $begingroup$

    Answer Should be




    075




    because




    0 Is right because there are two numbers right in first condition. And other one is wrong because of second condition in which 5 or 1 Has to be wrong. and From last condition 0 is selected and other 2 values got discard because of the 3rd condition. We got the position of the 0 from these two upper conditions. as 0 is wrong at second place in first condition and 0 is also wrong at last place in last condition.
    5 Is selected from first condition if we select 1 then we get only 2 numbers as we need 3 to make the key. 5 is also right in condition second in which we also got the right position of the 5.
    7 is right because of 4th condition as the other two number which is 1 is wrong due to first and second condition because we selected 5 and 6 is also wrong due to the position of 6. we need number that fits in position between 5 and 0. because 7 is at wrong place that fits at second position.
    As it satisfy all conditions
    501 — Two correct numbers in wrong places(5,0, As 1 or 5 get discard in second condition, so i took 5)
    135 — One correct number in the right place(5, As we can select only one that should be from first condition.)
    483 — All numbers are wrong(If presented above shall be discard.)
    167 — One correct number in wrong place(7, Both the last and first place is booked thats why only chance 7)
    430 — One correct number in the wrong place(0, as we took in the first step.)







    share|improve this answer











    $endgroup$





















      1












      $begingroup$

      Only the first four rules are necessary to find the answer.




      a. Via rules 2 & 3 the answer contains a 1 or a 5 but not both.

      b. Via (a), rule 1, and rule 2 the 1 or 5 in the answer is not in the middle.

      c. Rules 2 and 4 contradict if 1 is in the number, so 1 is not in the number, therefore 5 is at the end.

      d. Via (b), (c), and rule 1 the answer contains a 0 which is not in the middle, therefore 0 is at the front.

      e. Via (c), (d), and rule 4, 7 is in the middle.


      The answer is 075







      share|improve this answer









      $endgroup$





















        1












        $begingroup$

        I will share my process:




        • Notes of format !5 | !0 | !1 means that the first position can't be 5, second can't be 0, etc.


        • Possible is short for values which are possible and also have been previously discussed

        • Impossible represents values which have been ruled out

        • Definite represents values which have been confirmed to exist, though makes no statement on order

        • Conditions represent restrictions on password. For example: !(1+4) means the code cannot contain both one and 4


        Step one:




        !5 | !0 | !1

        Possible: 0,1,5
        Conditions: !(0+1+5)




        Step Two




        !3!5 | !0!1!5 | !1!3
        Possible: 0,1,3,5
        Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)




        Step Three




        !5 | !0!1!5 | !1
        Possible: 0,1,5
        Impossible: 3,4,8 -> as defined
        Conditions: !(0+1+5)




        Step Four




        !1!5 | !0!1!5!6 | !1!7 -> 1 is eliminated
        Possible: 0,5,6,7
        Impossible: 1,3,4,8
        Conditions: !(1+6+7)!(1+6)!(1+7)!(6+7)




        Step Five




        !5 | !0!5!6 | !7 -> !5 | 7 | !7 -> 6 is eliminated here, as explained in below:
        Possible: 0,5,6,7
        Impossible: 1,3,4,8
        Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed




        Conclusion




        Possible: 0,5,7
        First Term: !5 -> 0 (since 7 is second term)
        Second Term: 7
        Third Term: 5

        .
        0 7 5







        share|improve this answer











        $endgroup$





















          0












          $begingroup$

          1) 501

          2) 1X5 — 3 gets eliminated by the next clue

          3) XXX — All eliminated

          4) X67 — 1 gets eliminated for being in the "wrong" place here but the "right place" on Clue 2

          5) XX0 — 4 and 3 gets eliminated by clue 3, leaving 0




          So 0 is in the 1st position, because of the first clue that says it can't be in 2nd




          2) XX5 — The 0 took the 1's place, so only 5 is left




          So we have 0X5 at this point.




          4) XX7 — 1 gets eliminated by clues 1 and 2, and 6 would be in the "right" place if it was one of the numbers




          Answer: 075







          share|improve this answer











          $endgroup$





















            0












            $begingroup$


            Rule 1) 501 — Two correct numbers in wrong places

            Rule 2) 135 — One correct number in the right place

            Rule 3) 483 — All numbers are wrong

            Rule 4) 167 — One correct number in wrong place

            Rule 5) 430 — One correct number in the wrong place




            Starting from the most elimination Rule 3




            Because of Rule 3, Rule 5 proves that 0 is at position 1 or 2.

            But 0 can't be at position 2 because of Rule 1.


            So 0 is at position 1.




            Finding the 2nd digit




            Because of Rule 1, and we know about 0, it has to contain a 1 or a 5, but not both.

            And because of Rule 2, it can't be 1 because that position is already taken by 0.


            So 5 is at position 3.




            The remaining digit




            Then there's only the 2nd position to figure out.

            And because of Rule 4, it can't be 6.

            The 1 had already been discarded when we discovered 5 is at position 3.


            So 7 is at position 2.




            Conclusion:




            0 7 5







            share|improve this answer









            $endgroup$





















              -1












              $begingroup$

              The required password is




              075 .







              share|improve this answer











              $endgroup$













              • $begingroup$
                please explain your approach to this answer. thanks and happy puzzling ;)
                $endgroup$
                – Omega Krypton
                Feb 1 at 9:20











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              9 Answers
              9






              active

              oldest

              votes








              9 Answers
              9






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              11












              $begingroup$

              [5, 0, 1]: 2  
              [1, 3, 5]: 1



              So either 5 or 1 is correct, but not both.




              [4, 8, 3]: none.

              [1, 6, 7]: 1



              Because 1 is in the same place as it was before, but the "one correct" is in the wrong place, it cannot be a right number. This means that 5 (from earlier) is definitely the last number, and also one of either 6 or 7 is correct, but in the wrong place.




              [4, 3, 0]:



              We know from above that 4 and 3 are both wrong, and we deduced from the last step that 1 was not a number, meaning the first line shows us that 5 and 0 are both numbers, so this actually gives us no new information on which numbers, but does tell us that 0 is still in the wrong place.



              This means that we have "0 x 5" with x being either 6 or 7. From the second-last line, we know that the correct number is in the wrong place, so it can't be 6 (which was in the middle).



              The final answer is: 0 7 5.







              share|improve this answer











              $endgroup$













              • $begingroup$
                I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)
                $endgroup$
                – Kevin
                Jan 29 at 22:20
















              11












              $begingroup$

              [5, 0, 1]: 2  
              [1, 3, 5]: 1



              So either 5 or 1 is correct, but not both.




              [4, 8, 3]: none.

              [1, 6, 7]: 1



              Because 1 is in the same place as it was before, but the "one correct" is in the wrong place, it cannot be a right number. This means that 5 (from earlier) is definitely the last number, and also one of either 6 or 7 is correct, but in the wrong place.




              [4, 3, 0]:



              We know from above that 4 and 3 are both wrong, and we deduced from the last step that 1 was not a number, meaning the first line shows us that 5 and 0 are both numbers, so this actually gives us no new information on which numbers, but does tell us that 0 is still in the wrong place.



              This means that we have "0 x 5" with x being either 6 or 7. From the second-last line, we know that the correct number is in the wrong place, so it can't be 6 (which was in the middle).



              The final answer is: 0 7 5.







              share|improve this answer











              $endgroup$













              • $begingroup$
                I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)
                $endgroup$
                – Kevin
                Jan 29 at 22:20














              11












              11








              11





              $begingroup$

              [5, 0, 1]: 2  
              [1, 3, 5]: 1



              So either 5 or 1 is correct, but not both.




              [4, 8, 3]: none.

              [1, 6, 7]: 1



              Because 1 is in the same place as it was before, but the "one correct" is in the wrong place, it cannot be a right number. This means that 5 (from earlier) is definitely the last number, and also one of either 6 or 7 is correct, but in the wrong place.




              [4, 3, 0]:



              We know from above that 4 and 3 are both wrong, and we deduced from the last step that 1 was not a number, meaning the first line shows us that 5 and 0 are both numbers, so this actually gives us no new information on which numbers, but does tell us that 0 is still in the wrong place.



              This means that we have "0 x 5" with x being either 6 or 7. From the second-last line, we know that the correct number is in the wrong place, so it can't be 6 (which was in the middle).



              The final answer is: 0 7 5.







              share|improve this answer











              $endgroup$



              [5, 0, 1]: 2  
              [1, 3, 5]: 1



              So either 5 or 1 is correct, but not both.




              [4, 8, 3]: none.

              [1, 6, 7]: 1



              Because 1 is in the same place as it was before, but the "one correct" is in the wrong place, it cannot be a right number. This means that 5 (from earlier) is definitely the last number, and also one of either 6 or 7 is correct, but in the wrong place.




              [4, 3, 0]:



              We know from above that 4 and 3 are both wrong, and we deduced from the last step that 1 was not a number, meaning the first line shows us that 5 and 0 are both numbers, so this actually gives us no new information on which numbers, but does tell us that 0 is still in the wrong place.



              This means that we have "0 x 5" with x being either 6 or 7. From the second-last line, we know that the correct number is in the wrong place, so it can't be 6 (which was in the middle).



              The final answer is: 0 7 5.








              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Jan 29 at 16:55









              SteveV

              6,2502632




              6,2502632










              answered Jan 29 at 15:49









              Ian MacDonaldIan MacDonald

              11.2k2562




              11.2k2562












              • $begingroup$
                I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)
                $endgroup$
                – Kevin
                Jan 29 at 22:20


















              • $begingroup$
                I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)
                $endgroup$
                – Kevin
                Jan 29 at 22:20
















              $begingroup$
              I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)
              $endgroup$
              – Kevin
              Jan 29 at 22:20




              $begingroup$
              I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)
              $endgroup$
              – Kevin
              Jan 29 at 22:20











              3












              $begingroup$

              The following are my musings to find the answer:



              Starting with 430 (Rule 5):




              4?? Eliminated by rule 5
              ?4? Eliminated by rule 3
              ??4 Eliminated by rule 3
              3?? Eliminated by rule 3
              ?3? Eliminated by rule 5
              ??3 Eliminated by rule 3
              0?? Remaining - Now Rule 6
              ?0? Remaining - Now Rule 6
              ??0 Eliminated by rule 5




              Starting with 501 (Rule 1):




              5?1 Eliminated by rule 1
              51? Eliminated by rule 1
              15? Eliminated by rule 6
              1?5 Eliminated by rule 2
              ?15 Eliminated by rule 6
              ?51 Eliminated by rule 1
              50? Eliminated by rule 1
              5?0 Eliminated by rule 1
              ?50 Eliminated by rule 2
              ?05 Eliminated by rule 1
              0?5 Remaining - Now Rule 7
              05? Eliminated by rule 2
              ?01 Eliminated by rule 1
              ?10 Eliminated by rule 2
              1?0 Eliminated by rule 6
              10? Eliminated by rule 1
              01? Eliminated by rule 2
              0?1 Eliminated by rule 1




              Starting with 167 (Rule 4):




              1?? Elimated by rule 4
              ?1? Elimated by rule 1
              ??1 Elimated by rule 7
              6?? Elimated by rule 7
              ?6? Elimated by rule 4
              ??6 Elimated by rule 7
              7?? Elimated by rule 7
              ?7? Remaining - Now Rule 8
              ??7 Elimated by rule 4




              Combining the two new rules, 7 and 8 gives:




              Result 075







              share|improve this answer









              $endgroup$


















                3












                $begingroup$

                The following are my musings to find the answer:



                Starting with 430 (Rule 5):




                4?? Eliminated by rule 5
                ?4? Eliminated by rule 3
                ??4 Eliminated by rule 3
                3?? Eliminated by rule 3
                ?3? Eliminated by rule 5
                ??3 Eliminated by rule 3
                0?? Remaining - Now Rule 6
                ?0? Remaining - Now Rule 6
                ??0 Eliminated by rule 5




                Starting with 501 (Rule 1):




                5?1 Eliminated by rule 1
                51? Eliminated by rule 1
                15? Eliminated by rule 6
                1?5 Eliminated by rule 2
                ?15 Eliminated by rule 6
                ?51 Eliminated by rule 1
                50? Eliminated by rule 1
                5?0 Eliminated by rule 1
                ?50 Eliminated by rule 2
                ?05 Eliminated by rule 1
                0?5 Remaining - Now Rule 7
                05? Eliminated by rule 2
                ?01 Eliminated by rule 1
                ?10 Eliminated by rule 2
                1?0 Eliminated by rule 6
                10? Eliminated by rule 1
                01? Eliminated by rule 2
                0?1 Eliminated by rule 1




                Starting with 167 (Rule 4):




                1?? Elimated by rule 4
                ?1? Elimated by rule 1
                ??1 Elimated by rule 7
                6?? Elimated by rule 7
                ?6? Elimated by rule 4
                ??6 Elimated by rule 7
                7?? Elimated by rule 7
                ?7? Remaining - Now Rule 8
                ??7 Elimated by rule 4




                Combining the two new rules, 7 and 8 gives:




                Result 075







                share|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The following are my musings to find the answer:



                  Starting with 430 (Rule 5):




                  4?? Eliminated by rule 5
                  ?4? Eliminated by rule 3
                  ??4 Eliminated by rule 3
                  3?? Eliminated by rule 3
                  ?3? Eliminated by rule 5
                  ??3 Eliminated by rule 3
                  0?? Remaining - Now Rule 6
                  ?0? Remaining - Now Rule 6
                  ??0 Eliminated by rule 5




                  Starting with 501 (Rule 1):




                  5?1 Eliminated by rule 1
                  51? Eliminated by rule 1
                  15? Eliminated by rule 6
                  1?5 Eliminated by rule 2
                  ?15 Eliminated by rule 6
                  ?51 Eliminated by rule 1
                  50? Eliminated by rule 1
                  5?0 Eliminated by rule 1
                  ?50 Eliminated by rule 2
                  ?05 Eliminated by rule 1
                  0?5 Remaining - Now Rule 7
                  05? Eliminated by rule 2
                  ?01 Eliminated by rule 1
                  ?10 Eliminated by rule 2
                  1?0 Eliminated by rule 6
                  10? Eliminated by rule 1
                  01? Eliminated by rule 2
                  0?1 Eliminated by rule 1




                  Starting with 167 (Rule 4):




                  1?? Elimated by rule 4
                  ?1? Elimated by rule 1
                  ??1 Elimated by rule 7
                  6?? Elimated by rule 7
                  ?6? Elimated by rule 4
                  ??6 Elimated by rule 7
                  7?? Elimated by rule 7
                  ?7? Remaining - Now Rule 8
                  ??7 Elimated by rule 4




                  Combining the two new rules, 7 and 8 gives:




                  Result 075







                  share|improve this answer









                  $endgroup$



                  The following are my musings to find the answer:



                  Starting with 430 (Rule 5):




                  4?? Eliminated by rule 5
                  ?4? Eliminated by rule 3
                  ??4 Eliminated by rule 3
                  3?? Eliminated by rule 3
                  ?3? Eliminated by rule 5
                  ??3 Eliminated by rule 3
                  0?? Remaining - Now Rule 6
                  ?0? Remaining - Now Rule 6
                  ??0 Eliminated by rule 5




                  Starting with 501 (Rule 1):




                  5?1 Eliminated by rule 1
                  51? Eliminated by rule 1
                  15? Eliminated by rule 6
                  1?5 Eliminated by rule 2
                  ?15 Eliminated by rule 6
                  ?51 Eliminated by rule 1
                  50? Eliminated by rule 1
                  5?0 Eliminated by rule 1
                  ?50 Eliminated by rule 2
                  ?05 Eliminated by rule 1
                  0?5 Remaining - Now Rule 7
                  05? Eliminated by rule 2
                  ?01 Eliminated by rule 1
                  ?10 Eliminated by rule 2
                  1?0 Eliminated by rule 6
                  10? Eliminated by rule 1
                  01? Eliminated by rule 2
                  0?1 Eliminated by rule 1




                  Starting with 167 (Rule 4):




                  1?? Elimated by rule 4
                  ?1? Elimated by rule 1
                  ??1 Elimated by rule 7
                  6?? Elimated by rule 7
                  ?6? Elimated by rule 4
                  ??6 Elimated by rule 7
                  7?? Elimated by rule 7
                  ?7? Remaining - Now Rule 8
                  ??7 Elimated by rule 4




                  Combining the two new rules, 7 and 8 gives:




                  Result 075








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Jan 29 at 19:02









                  Tom CarpenterTom Carpenter

                  436210




                  436210























                      2












                      $begingroup$

                      I've found the answers too complicate, so I share mine. Only three rules are needed:



                      501 — Two correct numbers in wrong places




                      So, only one of those numbers should be removed. Which one?




                      135 — One correct number in the right place




                      Since this shares 2 numbers with the first one, the 3 cannot be the
                      correct one. (If the 3 comes into play, the rule would give info about
                      2 numbers instead of one). So, from the previous rule, we should
                      remove only the 1 or the 5. Now we know the number has a 0 in one side,
                      and starts with 1 or end with 5. Only 2 chances: 1_0 or 0_5.




                      167 — One correct number in wrong place




                      This number start with 1 as the previous one, and the 1 can't be in
                      right and wrong place at the same time. So, we have only one
                      possibility now: 0_5. We are only missing the middle number and can't be the
                      6 because it's already in the middle and not in the right place. So
                      the only option left is the 7. -> 075.







                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
                        $endgroup$
                        – North
                        Jan 30 at 3:06










                      • $begingroup$
                        Excellent observation about the 3 to get it down to three rules!
                        $endgroup$
                        – par
                        Jan 30 at 20:31


















                      2












                      $begingroup$

                      I've found the answers too complicate, so I share mine. Only three rules are needed:



                      501 — Two correct numbers in wrong places




                      So, only one of those numbers should be removed. Which one?




                      135 — One correct number in the right place




                      Since this shares 2 numbers with the first one, the 3 cannot be the
                      correct one. (If the 3 comes into play, the rule would give info about
                      2 numbers instead of one). So, from the previous rule, we should
                      remove only the 1 or the 5. Now we know the number has a 0 in one side,
                      and starts with 1 or end with 5. Only 2 chances: 1_0 or 0_5.




                      167 — One correct number in wrong place




                      This number start with 1 as the previous one, and the 1 can't be in
                      right and wrong place at the same time. So, we have only one
                      possibility now: 0_5. We are only missing the middle number and can't be the
                      6 because it's already in the middle and not in the right place. So
                      the only option left is the 7. -> 075.







                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
                        $endgroup$
                        – North
                        Jan 30 at 3:06










                      • $begingroup$
                        Excellent observation about the 3 to get it down to three rules!
                        $endgroup$
                        – par
                        Jan 30 at 20:31
















                      2












                      2








                      2





                      $begingroup$

                      I've found the answers too complicate, so I share mine. Only three rules are needed:



                      501 — Two correct numbers in wrong places




                      So, only one of those numbers should be removed. Which one?




                      135 — One correct number in the right place




                      Since this shares 2 numbers with the first one, the 3 cannot be the
                      correct one. (If the 3 comes into play, the rule would give info about
                      2 numbers instead of one). So, from the previous rule, we should
                      remove only the 1 or the 5. Now we know the number has a 0 in one side,
                      and starts with 1 or end with 5. Only 2 chances: 1_0 or 0_5.




                      167 — One correct number in wrong place




                      This number start with 1 as the previous one, and the 1 can't be in
                      right and wrong place at the same time. So, we have only one
                      possibility now: 0_5. We are only missing the middle number and can't be the
                      6 because it's already in the middle and not in the right place. So
                      the only option left is the 7. -> 075.







                      share|improve this answer









                      $endgroup$



                      I've found the answers too complicate, so I share mine. Only three rules are needed:



                      501 — Two correct numbers in wrong places




                      So, only one of those numbers should be removed. Which one?




                      135 — One correct number in the right place




                      Since this shares 2 numbers with the first one, the 3 cannot be the
                      correct one. (If the 3 comes into play, the rule would give info about
                      2 numbers instead of one). So, from the previous rule, we should
                      remove only the 1 or the 5. Now we know the number has a 0 in one side,
                      and starts with 1 or end with 5. Only 2 chances: 1_0 or 0_5.




                      167 — One correct number in wrong place




                      This number start with 1 as the previous one, and the 1 can't be in
                      right and wrong place at the same time. So, we have only one
                      possibility now: 0_5. We are only missing the middle number and can't be the
                      6 because it's already in the middle and not in the right place. So
                      the only option left is the 7. -> 075.








                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Jan 30 at 2:54









                      TooLazyToLogInMyAccTooLazyToLogInMyAcc

                      211




                      211












                      • $begingroup$
                        Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
                        $endgroup$
                        – North
                        Jan 30 at 3:06










                      • $begingroup$
                        Excellent observation about the 3 to get it down to three rules!
                        $endgroup$
                        – par
                        Jan 30 at 20:31




















                      • $begingroup$
                        Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
                        $endgroup$
                        – North
                        Jan 30 at 3:06










                      • $begingroup$
                        Excellent observation about the 3 to get it down to three rules!
                        $endgroup$
                        – par
                        Jan 30 at 20:31


















                      $begingroup$
                      Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
                      $endgroup$
                      – North
                      Jan 30 at 3:06




                      $begingroup$
                      Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
                      $endgroup$
                      – North
                      Jan 30 at 3:06












                      $begingroup$
                      Excellent observation about the 3 to get it down to three rules!
                      $endgroup$
                      – par
                      Jan 30 at 20:31






                      $begingroup$
                      Excellent observation about the 3 to get it down to three rules!
                      $endgroup$
                      – par
                      Jan 30 at 20:31













                      2












                      $begingroup$

                      Answer Should be




                      075




                      because




                      0 Is right because there are two numbers right in first condition. And other one is wrong because of second condition in which 5 or 1 Has to be wrong. and From last condition 0 is selected and other 2 values got discard because of the 3rd condition. We got the position of the 0 from these two upper conditions. as 0 is wrong at second place in first condition and 0 is also wrong at last place in last condition.
                      5 Is selected from first condition if we select 1 then we get only 2 numbers as we need 3 to make the key. 5 is also right in condition second in which we also got the right position of the 5.
                      7 is right because of 4th condition as the other two number which is 1 is wrong due to first and second condition because we selected 5 and 6 is also wrong due to the position of 6. we need number that fits in position between 5 and 0. because 7 is at wrong place that fits at second position.
                      As it satisfy all conditions
                      501 — Two correct numbers in wrong places(5,0, As 1 or 5 get discard in second condition, so i took 5)
                      135 — One correct number in the right place(5, As we can select only one that should be from first condition.)
                      483 — All numbers are wrong(If presented above shall be discard.)
                      167 — One correct number in wrong place(7, Both the last and first place is booked thats why only chance 7)
                      430 — One correct number in the wrong place(0, as we took in the first step.)







                      share|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        Answer Should be




                        075




                        because




                        0 Is right because there are two numbers right in first condition. And other one is wrong because of second condition in which 5 or 1 Has to be wrong. and From last condition 0 is selected and other 2 values got discard because of the 3rd condition. We got the position of the 0 from these two upper conditions. as 0 is wrong at second place in first condition and 0 is also wrong at last place in last condition.
                        5 Is selected from first condition if we select 1 then we get only 2 numbers as we need 3 to make the key. 5 is also right in condition second in which we also got the right position of the 5.
                        7 is right because of 4th condition as the other two number which is 1 is wrong due to first and second condition because we selected 5 and 6 is also wrong due to the position of 6. we need number that fits in position between 5 and 0. because 7 is at wrong place that fits at second position.
                        As it satisfy all conditions
                        501 — Two correct numbers in wrong places(5,0, As 1 or 5 get discard in second condition, so i took 5)
                        135 — One correct number in the right place(5, As we can select only one that should be from first condition.)
                        483 — All numbers are wrong(If presented above shall be discard.)
                        167 — One correct number in wrong place(7, Both the last and first place is booked thats why only chance 7)
                        430 — One correct number in the wrong place(0, as we took in the first step.)







                        share|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Answer Should be




                          075




                          because




                          0 Is right because there are two numbers right in first condition. And other one is wrong because of second condition in which 5 or 1 Has to be wrong. and From last condition 0 is selected and other 2 values got discard because of the 3rd condition. We got the position of the 0 from these two upper conditions. as 0 is wrong at second place in first condition and 0 is also wrong at last place in last condition.
                          5 Is selected from first condition if we select 1 then we get only 2 numbers as we need 3 to make the key. 5 is also right in condition second in which we also got the right position of the 5.
                          7 is right because of 4th condition as the other two number which is 1 is wrong due to first and second condition because we selected 5 and 6 is also wrong due to the position of 6. we need number that fits in position between 5 and 0. because 7 is at wrong place that fits at second position.
                          As it satisfy all conditions
                          501 — Two correct numbers in wrong places(5,0, As 1 or 5 get discard in second condition, so i took 5)
                          135 — One correct number in the right place(5, As we can select only one that should be from first condition.)
                          483 — All numbers are wrong(If presented above shall be discard.)
                          167 — One correct number in wrong place(7, Both the last and first place is booked thats why only chance 7)
                          430 — One correct number in the wrong place(0, as we took in the first step.)







                          share|improve this answer











                          $endgroup$



                          Answer Should be




                          075




                          because




                          0 Is right because there are two numbers right in first condition. And other one is wrong because of second condition in which 5 or 1 Has to be wrong. and From last condition 0 is selected and other 2 values got discard because of the 3rd condition. We got the position of the 0 from these two upper conditions. as 0 is wrong at second place in first condition and 0 is also wrong at last place in last condition.
                          5 Is selected from first condition if we select 1 then we get only 2 numbers as we need 3 to make the key. 5 is also right in condition second in which we also got the right position of the 5.
                          7 is right because of 4th condition as the other two number which is 1 is wrong due to first and second condition because we selected 5 and 6 is also wrong due to the position of 6. we need number that fits in position between 5 and 0. because 7 is at wrong place that fits at second position.
                          As it satisfy all conditions
                          501 — Two correct numbers in wrong places(5,0, As 1 or 5 get discard in second condition, so i took 5)
                          135 — One correct number in the right place(5, As we can select only one that should be from first condition.)
                          483 — All numbers are wrong(If presented above shall be discard.)
                          167 — One correct number in wrong place(7, Both the last and first place is booked thats why only chance 7)
                          430 — One correct number in the wrong place(0, as we took in the first step.)








                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Jan 30 at 9:51









                          Flying_whale

                          1,909425




                          1,909425










                          answered Jan 29 at 17:05









                          CrackItCrackIt

                          212




                          212























                              1












                              $begingroup$

                              Only the first four rules are necessary to find the answer.




                              a. Via rules 2 & 3 the answer contains a 1 or a 5 but not both.

                              b. Via (a), rule 1, and rule 2 the 1 or 5 in the answer is not in the middle.

                              c. Rules 2 and 4 contradict if 1 is in the number, so 1 is not in the number, therefore 5 is at the end.

                              d. Via (b), (c), and rule 1 the answer contains a 0 which is not in the middle, therefore 0 is at the front.

                              e. Via (c), (d), and rule 4, 7 is in the middle.


                              The answer is 075







                              share|improve this answer









                              $endgroup$


















                                1












                                $begingroup$

                                Only the first four rules are necessary to find the answer.




                                a. Via rules 2 & 3 the answer contains a 1 or a 5 but not both.

                                b. Via (a), rule 1, and rule 2 the 1 or 5 in the answer is not in the middle.

                                c. Rules 2 and 4 contradict if 1 is in the number, so 1 is not in the number, therefore 5 is at the end.

                                d. Via (b), (c), and rule 1 the answer contains a 0 which is not in the middle, therefore 0 is at the front.

                                e. Via (c), (d), and rule 4, 7 is in the middle.


                                The answer is 075







                                share|improve this answer









                                $endgroup$
















                                  1












                                  1








                                  1





                                  $begingroup$

                                  Only the first four rules are necessary to find the answer.




                                  a. Via rules 2 & 3 the answer contains a 1 or a 5 but not both.

                                  b. Via (a), rule 1, and rule 2 the 1 or 5 in the answer is not in the middle.

                                  c. Rules 2 and 4 contradict if 1 is in the number, so 1 is not in the number, therefore 5 is at the end.

                                  d. Via (b), (c), and rule 1 the answer contains a 0 which is not in the middle, therefore 0 is at the front.

                                  e. Via (c), (d), and rule 4, 7 is in the middle.


                                  The answer is 075







                                  share|improve this answer









                                  $endgroup$



                                  Only the first four rules are necessary to find the answer.




                                  a. Via rules 2 & 3 the answer contains a 1 or a 5 but not both.

                                  b. Via (a), rule 1, and rule 2 the 1 or 5 in the answer is not in the middle.

                                  c. Rules 2 and 4 contradict if 1 is in the number, so 1 is not in the number, therefore 5 is at the end.

                                  d. Via (b), (c), and rule 1 the answer contains a 0 which is not in the middle, therefore 0 is at the front.

                                  e. Via (c), (d), and rule 4, 7 is in the middle.


                                  The answer is 075








                                  share|improve this answer












                                  share|improve this answer



                                  share|improve this answer










                                  answered Jan 29 at 20:10









                                  parpar

                                  1113




                                  1113























                                      1












                                      $begingroup$

                                      I will share my process:




                                      • Notes of format !5 | !0 | !1 means that the first position can't be 5, second can't be 0, etc.


                                      • Possible is short for values which are possible and also have been previously discussed

                                      • Impossible represents values which have been ruled out

                                      • Definite represents values which have been confirmed to exist, though makes no statement on order

                                      • Conditions represent restrictions on password. For example: !(1+4) means the code cannot contain both one and 4


                                      Step one:




                                      !5 | !0 | !1

                                      Possible: 0,1,5
                                      Conditions: !(0+1+5)




                                      Step Two




                                      !3!5 | !0!1!5 | !1!3
                                      Possible: 0,1,3,5
                                      Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)




                                      Step Three




                                      !5 | !0!1!5 | !1
                                      Possible: 0,1,5
                                      Impossible: 3,4,8 -> as defined
                                      Conditions: !(0+1+5)




                                      Step Four




                                      !1!5 | !0!1!5!6 | !1!7 -> 1 is eliminated
                                      Possible: 0,5,6,7
                                      Impossible: 1,3,4,8
                                      Conditions: !(1+6+7)!(1+6)!(1+7)!(6+7)




                                      Step Five




                                      !5 | !0!5!6 | !7 -> !5 | 7 | !7 -> 6 is eliminated here, as explained in below:
                                      Possible: 0,5,6,7
                                      Impossible: 1,3,4,8
                                      Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed




                                      Conclusion




                                      Possible: 0,5,7
                                      First Term: !5 -> 0 (since 7 is second term)
                                      Second Term: 7
                                      Third Term: 5

                                      .
                                      0 7 5







                                      share|improve this answer











                                      $endgroup$


















                                        1












                                        $begingroup$

                                        I will share my process:




                                        • Notes of format !5 | !0 | !1 means that the first position can't be 5, second can't be 0, etc.


                                        • Possible is short for values which are possible and also have been previously discussed

                                        • Impossible represents values which have been ruled out

                                        • Definite represents values which have been confirmed to exist, though makes no statement on order

                                        • Conditions represent restrictions on password. For example: !(1+4) means the code cannot contain both one and 4


                                        Step one:




                                        !5 | !0 | !1

                                        Possible: 0,1,5
                                        Conditions: !(0+1+5)




                                        Step Two




                                        !3!5 | !0!1!5 | !1!3
                                        Possible: 0,1,3,5
                                        Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)




                                        Step Three




                                        !5 | !0!1!5 | !1
                                        Possible: 0,1,5
                                        Impossible: 3,4,8 -> as defined
                                        Conditions: !(0+1+5)




                                        Step Four




                                        !1!5 | !0!1!5!6 | !1!7 -> 1 is eliminated
                                        Possible: 0,5,6,7
                                        Impossible: 1,3,4,8
                                        Conditions: !(1+6+7)!(1+6)!(1+7)!(6+7)




                                        Step Five




                                        !5 | !0!5!6 | !7 -> !5 | 7 | !7 -> 6 is eliminated here, as explained in below:
                                        Possible: 0,5,6,7
                                        Impossible: 1,3,4,8
                                        Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed




                                        Conclusion




                                        Possible: 0,5,7
                                        First Term: !5 -> 0 (since 7 is second term)
                                        Second Term: 7
                                        Third Term: 5

                                        .
                                        0 7 5







                                        share|improve this answer











                                        $endgroup$
















                                          1












                                          1








                                          1





                                          $begingroup$

                                          I will share my process:




                                          • Notes of format !5 | !0 | !1 means that the first position can't be 5, second can't be 0, etc.


                                          • Possible is short for values which are possible and also have been previously discussed

                                          • Impossible represents values which have been ruled out

                                          • Definite represents values which have been confirmed to exist, though makes no statement on order

                                          • Conditions represent restrictions on password. For example: !(1+4) means the code cannot contain both one and 4


                                          Step one:




                                          !5 | !0 | !1

                                          Possible: 0,1,5
                                          Conditions: !(0+1+5)




                                          Step Two




                                          !3!5 | !0!1!5 | !1!3
                                          Possible: 0,1,3,5
                                          Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)




                                          Step Three




                                          !5 | !0!1!5 | !1
                                          Possible: 0,1,5
                                          Impossible: 3,4,8 -> as defined
                                          Conditions: !(0+1+5)




                                          Step Four




                                          !1!5 | !0!1!5!6 | !1!7 -> 1 is eliminated
                                          Possible: 0,5,6,7
                                          Impossible: 1,3,4,8
                                          Conditions: !(1+6+7)!(1+6)!(1+7)!(6+7)




                                          Step Five




                                          !5 | !0!5!6 | !7 -> !5 | 7 | !7 -> 6 is eliminated here, as explained in below:
                                          Possible: 0,5,6,7
                                          Impossible: 1,3,4,8
                                          Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed




                                          Conclusion




                                          Possible: 0,5,7
                                          First Term: !5 -> 0 (since 7 is second term)
                                          Second Term: 7
                                          Third Term: 5

                                          .
                                          0 7 5







                                          share|improve this answer











                                          $endgroup$



                                          I will share my process:




                                          • Notes of format !5 | !0 | !1 means that the first position can't be 5, second can't be 0, etc.


                                          • Possible is short for values which are possible and also have been previously discussed

                                          • Impossible represents values which have been ruled out

                                          • Definite represents values which have been confirmed to exist, though makes no statement on order

                                          • Conditions represent restrictions on password. For example: !(1+4) means the code cannot contain both one and 4


                                          Step one:




                                          !5 | !0 | !1

                                          Possible: 0,1,5
                                          Conditions: !(0+1+5)




                                          Step Two




                                          !3!5 | !0!1!5 | !1!3
                                          Possible: 0,1,3,5
                                          Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)




                                          Step Three




                                          !5 | !0!1!5 | !1
                                          Possible: 0,1,5
                                          Impossible: 3,4,8 -> as defined
                                          Conditions: !(0+1+5)




                                          Step Four




                                          !1!5 | !0!1!5!6 | !1!7 -> 1 is eliminated
                                          Possible: 0,5,6,7
                                          Impossible: 1,3,4,8
                                          Conditions: !(1+6+7)!(1+6)!(1+7)!(6+7)




                                          Step Five




                                          !5 | !0!5!6 | !7 -> !5 | 7 | !7 -> 6 is eliminated here, as explained in below:
                                          Possible: 0,5,6,7
                                          Impossible: 1,3,4,8
                                          Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed




                                          Conclusion




                                          Possible: 0,5,7
                                          First Term: !5 -> 0 (since 7 is second term)
                                          Second Term: 7
                                          Third Term: 5

                                          .
                                          0 7 5








                                          share|improve this answer














                                          share|improve this answer



                                          share|improve this answer








                                          edited Jan 30 at 4:13

























                                          answered Jan 29 at 20:22









                                          Brandon DixonBrandon Dixon

                                          1212




                                          1212























                                              0












                                              $begingroup$

                                              1) 501

                                              2) 1X5 — 3 gets eliminated by the next clue

                                              3) XXX — All eliminated

                                              4) X67 — 1 gets eliminated for being in the "wrong" place here but the "right place" on Clue 2

                                              5) XX0 — 4 and 3 gets eliminated by clue 3, leaving 0




                                              So 0 is in the 1st position, because of the first clue that says it can't be in 2nd




                                              2) XX5 — The 0 took the 1's place, so only 5 is left




                                              So we have 0X5 at this point.




                                              4) XX7 — 1 gets eliminated by clues 1 and 2, and 6 would be in the "right" place if it was one of the numbers




                                              Answer: 075







                                              share|improve this answer











                                              $endgroup$


















                                                0












                                                $begingroup$

                                                1) 501

                                                2) 1X5 — 3 gets eliminated by the next clue

                                                3) XXX — All eliminated

                                                4) X67 — 1 gets eliminated for being in the "wrong" place here but the "right place" on Clue 2

                                                5) XX0 — 4 and 3 gets eliminated by clue 3, leaving 0




                                                So 0 is in the 1st position, because of the first clue that says it can't be in 2nd




                                                2) XX5 — The 0 took the 1's place, so only 5 is left




                                                So we have 0X5 at this point.




                                                4) XX7 — 1 gets eliminated by clues 1 and 2, and 6 would be in the "right" place if it was one of the numbers




                                                Answer: 075







                                                share|improve this answer











                                                $endgroup$
















                                                  0












                                                  0








                                                  0





                                                  $begingroup$

                                                  1) 501

                                                  2) 1X5 — 3 gets eliminated by the next clue

                                                  3) XXX — All eliminated

                                                  4) X67 — 1 gets eliminated for being in the "wrong" place here but the "right place" on Clue 2

                                                  5) XX0 — 4 and 3 gets eliminated by clue 3, leaving 0




                                                  So 0 is in the 1st position, because of the first clue that says it can't be in 2nd




                                                  2) XX5 — The 0 took the 1's place, so only 5 is left




                                                  So we have 0X5 at this point.




                                                  4) XX7 — 1 gets eliminated by clues 1 and 2, and 6 would be in the "right" place if it was one of the numbers




                                                  Answer: 075







                                                  share|improve this answer











                                                  $endgroup$



                                                  1) 501

                                                  2) 1X5 — 3 gets eliminated by the next clue

                                                  3) XXX — All eliminated

                                                  4) X67 — 1 gets eliminated for being in the "wrong" place here but the "right place" on Clue 2

                                                  5) XX0 — 4 and 3 gets eliminated by clue 3, leaving 0




                                                  So 0 is in the 1st position, because of the first clue that says it can't be in 2nd




                                                  2) XX5 — The 0 took the 1's place, so only 5 is left




                                                  So we have 0X5 at this point.




                                                  4) XX7 — 1 gets eliminated by clues 1 and 2, and 6 would be in the "right" place if it was one of the numbers




                                                  Answer: 075








                                                  share|improve this answer














                                                  share|improve this answer



                                                  share|improve this answer








                                                  edited Jan 29 at 19:10

























                                                  answered Jan 29 at 19:05









                                                  J.-S. PayetteJ.-S. Payette

                                                  213




                                                  213























                                                      0












                                                      $begingroup$


                                                      Rule 1) 501 — Two correct numbers in wrong places

                                                      Rule 2) 135 — One correct number in the right place

                                                      Rule 3) 483 — All numbers are wrong

                                                      Rule 4) 167 — One correct number in wrong place

                                                      Rule 5) 430 — One correct number in the wrong place




                                                      Starting from the most elimination Rule 3




                                                      Because of Rule 3, Rule 5 proves that 0 is at position 1 or 2.

                                                      But 0 can't be at position 2 because of Rule 1.


                                                      So 0 is at position 1.




                                                      Finding the 2nd digit




                                                      Because of Rule 1, and we know about 0, it has to contain a 1 or a 5, but not both.

                                                      And because of Rule 2, it can't be 1 because that position is already taken by 0.


                                                      So 5 is at position 3.




                                                      The remaining digit




                                                      Then there's only the 2nd position to figure out.

                                                      And because of Rule 4, it can't be 6.

                                                      The 1 had already been discarded when we discovered 5 is at position 3.


                                                      So 7 is at position 2.




                                                      Conclusion:




                                                      0 7 5







                                                      share|improve this answer









                                                      $endgroup$


















                                                        0












                                                        $begingroup$


                                                        Rule 1) 501 — Two correct numbers in wrong places

                                                        Rule 2) 135 — One correct number in the right place

                                                        Rule 3) 483 — All numbers are wrong

                                                        Rule 4) 167 — One correct number in wrong place

                                                        Rule 5) 430 — One correct number in the wrong place




                                                        Starting from the most elimination Rule 3




                                                        Because of Rule 3, Rule 5 proves that 0 is at position 1 or 2.

                                                        But 0 can't be at position 2 because of Rule 1.


                                                        So 0 is at position 1.




                                                        Finding the 2nd digit




                                                        Because of Rule 1, and we know about 0, it has to contain a 1 or a 5, but not both.

                                                        And because of Rule 2, it can't be 1 because that position is already taken by 0.


                                                        So 5 is at position 3.




                                                        The remaining digit




                                                        Then there's only the 2nd position to figure out.

                                                        And because of Rule 4, it can't be 6.

                                                        The 1 had already been discarded when we discovered 5 is at position 3.


                                                        So 7 is at position 2.




                                                        Conclusion:




                                                        0 7 5







                                                        share|improve this answer









                                                        $endgroup$
















                                                          0












                                                          0








                                                          0





                                                          $begingroup$


                                                          Rule 1) 501 — Two correct numbers in wrong places

                                                          Rule 2) 135 — One correct number in the right place

                                                          Rule 3) 483 — All numbers are wrong

                                                          Rule 4) 167 — One correct number in wrong place

                                                          Rule 5) 430 — One correct number in the wrong place




                                                          Starting from the most elimination Rule 3




                                                          Because of Rule 3, Rule 5 proves that 0 is at position 1 or 2.

                                                          But 0 can't be at position 2 because of Rule 1.


                                                          So 0 is at position 1.




                                                          Finding the 2nd digit




                                                          Because of Rule 1, and we know about 0, it has to contain a 1 or a 5, but not both.

                                                          And because of Rule 2, it can't be 1 because that position is already taken by 0.


                                                          So 5 is at position 3.




                                                          The remaining digit




                                                          Then there's only the 2nd position to figure out.

                                                          And because of Rule 4, it can't be 6.

                                                          The 1 had already been discarded when we discovered 5 is at position 3.


                                                          So 7 is at position 2.




                                                          Conclusion:




                                                          0 7 5







                                                          share|improve this answer









                                                          $endgroup$




                                                          Rule 1) 501 — Two correct numbers in wrong places

                                                          Rule 2) 135 — One correct number in the right place

                                                          Rule 3) 483 — All numbers are wrong

                                                          Rule 4) 167 — One correct number in wrong place

                                                          Rule 5) 430 — One correct number in the wrong place




                                                          Starting from the most elimination Rule 3




                                                          Because of Rule 3, Rule 5 proves that 0 is at position 1 or 2.

                                                          But 0 can't be at position 2 because of Rule 1.


                                                          So 0 is at position 1.




                                                          Finding the 2nd digit




                                                          Because of Rule 1, and we know about 0, it has to contain a 1 or a 5, but not both.

                                                          And because of Rule 2, it can't be 1 because that position is already taken by 0.


                                                          So 5 is at position 3.




                                                          The remaining digit




                                                          Then there's only the 2nd position to figure out.

                                                          And because of Rule 4, it can't be 6.

                                                          The 1 had already been discarded when we discovered 5 is at position 3.


                                                          So 7 is at position 2.




                                                          Conclusion:




                                                          0 7 5








                                                          share|improve this answer












                                                          share|improve this answer



                                                          share|improve this answer










                                                          answered Jan 30 at 13:24









                                                          LukStormsLukStorms

                                                          1013




                                                          1013























                                                              -1












                                                              $begingroup$

                                                              The required password is




                                                              075 .







                                                              share|improve this answer











                                                              $endgroup$













                                                              • $begingroup$
                                                                please explain your approach to this answer. thanks and happy puzzling ;)
                                                                $endgroup$
                                                                – Omega Krypton
                                                                Feb 1 at 9:20
















                                                              -1












                                                              $begingroup$

                                                              The required password is




                                                              075 .







                                                              share|improve this answer











                                                              $endgroup$













                                                              • $begingroup$
                                                                please explain your approach to this answer. thanks and happy puzzling ;)
                                                                $endgroup$
                                                                – Omega Krypton
                                                                Feb 1 at 9:20














                                                              -1












                                                              -1








                                                              -1





                                                              $begingroup$

                                                              The required password is




                                                              075 .







                                                              share|improve this answer











                                                              $endgroup$



                                                              The required password is




                                                              075 .








                                                              share|improve this answer














                                                              share|improve this answer



                                                              share|improve this answer








                                                              edited Feb 1 at 8:32









                                                              Dmihawk

                                                              2,493828




                                                              2,493828










                                                              answered Feb 1 at 8:24









                                                              Manish MasiwalManish Masiwal

                                                              1




                                                              1












                                                              • $begingroup$
                                                                please explain your approach to this answer. thanks and happy puzzling ;)
                                                                $endgroup$
                                                                – Omega Krypton
                                                                Feb 1 at 9:20


















                                                              • $begingroup$
                                                                please explain your approach to this answer. thanks and happy puzzling ;)
                                                                $endgroup$
                                                                – Omega Krypton
                                                                Feb 1 at 9:20
















                                                              $begingroup$
                                                              please explain your approach to this answer. thanks and happy puzzling ;)
                                                              $endgroup$
                                                              – Omega Krypton
                                                              Feb 1 at 9:20




                                                              $begingroup$
                                                              please explain your approach to this answer. thanks and happy puzzling ;)
                                                              $endgroup$
                                                              – Omega Krypton
                                                              Feb 1 at 9:20


















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