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Adding up numbers in Portuguese is strange


How does one go about picking words when creating Verbal Arithmetic puzzles?SPEND LESS MONEY alphameticHow do I solve cryptarithmetic puzzles?Letters = numbersNumbers = letters 2.0Does AAA+BBB+CCC+DDD=ABCD have a solution for distinct digits A,B,C,D?Arrange numbers to 3 different math operationsSum of three six digit numbersWhat are these numbers?













9












$begingroup$


My friend from Portugal said that 2 + 2 is 8 in Portuguese. And he could prove it. I did not understand it at first. But then he showed me the following equation:



 D O I S
D O I S
+ ________
O I T O


Ah, now it is clear. Two is "dois" and eight is "oito" in Portuguese. And the equation above makes perfect sense. Could you solve the single one solution that explain this?



Rules: D, O, I, S and T are a number between 0 - 9. Each one is unique.










share|improve this question









$endgroup$
















    9












    $begingroup$


    My friend from Portugal said that 2 + 2 is 8 in Portuguese. And he could prove it. I did not understand it at first. But then he showed me the following equation:



     D O I S
    D O I S
    + ________
    O I T O


    Ah, now it is clear. Two is "dois" and eight is "oito" in Portuguese. And the equation above makes perfect sense. Could you solve the single one solution that explain this?



    Rules: D, O, I, S and T are a number between 0 - 9. Each one is unique.










    share|improve this question









    $endgroup$














      9












      9








      9





      $begingroup$


      My friend from Portugal said that 2 + 2 is 8 in Portuguese. And he could prove it. I did not understand it at first. But then he showed me the following equation:



       D O I S
      D O I S
      + ________
      O I T O


      Ah, now it is clear. Two is "dois" and eight is "oito" in Portuguese. And the equation above makes perfect sense. Could you solve the single one solution that explain this?



      Rules: D, O, I, S and T are a number between 0 - 9. Each one is unique.










      share|improve this question









      $endgroup$




      My friend from Portugal said that 2 + 2 is 8 in Portuguese. And he could prove it. I did not understand it at first. But then he showed me the following equation:



       D O I S
      D O I S
      + ________
      O I T O


      Ah, now it is clear. Two is "dois" and eight is "oito" in Portuguese. And the equation above makes perfect sense. Could you solve the single one solution that explain this?



      Rules: D, O, I, S and T are a number between 0 - 9. Each one is unique.







      alphametic






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 20 at 4:09









      ChaoticChaotic

      621316




      621316




















          3 Answers
          3






          active

          oldest

          votes


















          10












          $begingroup$

          Could it be





           1246 

          + 1246

          ------

          2492



          Then




          D=1, O=2, I=4, S=6, T=9







          share|improve this answer











          $endgroup$












          • $begingroup$
            Perfect! Easy one?
            $endgroup$
            – Chaotic
            Mar 20 at 4:50






          • 1




            $begingroup$
            Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
            $endgroup$
            – El-Guest
            Mar 20 at 11:28


















          13












          $begingroup$


          From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.







          share|improve this answer











          $endgroup$












          • $begingroup$
            You haven't given a value for $T$.
            $endgroup$
            – ZanyG
            Mar 20 at 5:54


















          0












          $begingroup$

          Let's start as a basic Brute Force





          Consider D as 1

          Since D and S gives a Unit Place as O

          So,S is D+5 ;S is 6

          So, Here O is 2

          Given I = O + O; So,I=4

          Since T is I+I with a carry of 1 ;T is 9






          share|improve this answer











          $endgroup$













            Your Answer








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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            Could it be





             1246 

            + 1246

            ------

            2492



            Then




            D=1, O=2, I=4, S=6, T=9







            share|improve this answer











            $endgroup$












            • $begingroup$
              Perfect! Easy one?
              $endgroup$
              – Chaotic
              Mar 20 at 4:50






            • 1




              $begingroup$
              Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
              $endgroup$
              – El-Guest
              Mar 20 at 11:28















            10












            $begingroup$

            Could it be





             1246 

            + 1246

            ------

            2492



            Then




            D=1, O=2, I=4, S=6, T=9







            share|improve this answer











            $endgroup$












            • $begingroup$
              Perfect! Easy one?
              $endgroup$
              – Chaotic
              Mar 20 at 4:50






            • 1




              $begingroup$
              Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
              $endgroup$
              – El-Guest
              Mar 20 at 11:28













            10












            10








            10





            $begingroup$

            Could it be





             1246 

            + 1246

            ------

            2492



            Then




            D=1, O=2, I=4, S=6, T=9







            share|improve this answer











            $endgroup$



            Could it be





             1246 

            + 1246

            ------

            2492



            Then




            D=1, O=2, I=4, S=6, T=9








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 20 at 12:40









            Omega Krypton

            5,5892849




            5,5892849










            answered Mar 20 at 4:42









            El-GuestEl-Guest

            22.2k35193




            22.2k35193











            • $begingroup$
              Perfect! Easy one?
              $endgroup$
              – Chaotic
              Mar 20 at 4:50






            • 1




              $begingroup$
              Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
              $endgroup$
              – El-Guest
              Mar 20 at 11:28
















            • $begingroup$
              Perfect! Easy one?
              $endgroup$
              – Chaotic
              Mar 20 at 4:50






            • 1




              $begingroup$
              Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
              $endgroup$
              – El-Guest
              Mar 20 at 11:28















            $begingroup$
            Perfect! Easy one?
            $endgroup$
            – Chaotic
            Mar 20 at 4:50




            $begingroup$
            Perfect! Easy one?
            $endgroup$
            – Chaotic
            Mar 20 at 4:50




            1




            1




            $begingroup$
            Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
            $endgroup$
            – El-Guest
            Mar 20 at 11:28




            $begingroup$
            Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
            $endgroup$
            – El-Guest
            Mar 20 at 11:28











            13












            $begingroup$


            From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.







            share|improve this answer











            $endgroup$












            • $begingroup$
              You haven't given a value for $T$.
              $endgroup$
              – ZanyG
              Mar 20 at 5:54















            13












            $begingroup$


            From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.







            share|improve this answer











            $endgroup$












            • $begingroup$
              You haven't given a value for $T$.
              $endgroup$
              – ZanyG
              Mar 20 at 5:54













            13












            13








            13





            $begingroup$


            From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.







            share|improve this answer











            $endgroup$




            From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 20 at 9:03

























            answered Mar 20 at 5:16









            JonMark PerryJonMark Perry

            21.1k641100




            21.1k641100











            • $begingroup$
              You haven't given a value for $T$.
              $endgroup$
              – ZanyG
              Mar 20 at 5:54
















            • $begingroup$
              You haven't given a value for $T$.
              $endgroup$
              – ZanyG
              Mar 20 at 5:54















            $begingroup$
            You haven't given a value for $T$.
            $endgroup$
            – ZanyG
            Mar 20 at 5:54




            $begingroup$
            You haven't given a value for $T$.
            $endgroup$
            – ZanyG
            Mar 20 at 5:54











            0












            $begingroup$

            Let's start as a basic Brute Force





            Consider D as 1

            Since D and S gives a Unit Place as O

            So,S is D+5 ;S is 6

            So, Here O is 2

            Given I = O + O; So,I=4

            Since T is I+I with a carry of 1 ;T is 9






            share|improve this answer











            $endgroup$

















              0












              $begingroup$

              Let's start as a basic Brute Force





              Consider D as 1

              Since D and S gives a Unit Place as O

              So,S is D+5 ;S is 6

              So, Here O is 2

              Given I = O + O; So,I=4

              Since T is I+I with a carry of 1 ;T is 9






              share|improve this answer











              $endgroup$















                0












                0








                0





                $begingroup$

                Let's start as a basic Brute Force





                Consider D as 1

                Since D and S gives a Unit Place as O

                So,S is D+5 ;S is 6

                So, Here O is 2

                Given I = O + O; So,I=4

                Since T is I+I with a carry of 1 ;T is 9






                share|improve this answer











                $endgroup$



                Let's start as a basic Brute Force





                Consider D as 1

                Since D and S gives a Unit Place as O

                So,S is D+5 ;S is 6

                So, Here O is 2

                Given I = O + O; So,I=4

                Since T is I+I with a carry of 1 ;T is 9







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Mar 20 at 8:15









                JonMark Perry

                21.1k641100




                21.1k641100










                answered Mar 20 at 6:12









                NaveenGopal NolluNaveenGopal Nollu

                1




                1



























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