how can it be possible to add three odd numbers and the answer is even












5












$begingroup$


From the days I started to learn Maths, I've have been taught that




Adding Odd times Odd numbers the Answer always would be Odd

e.g 3 + 5 + 1 = 9




OK, but look at this question



UPSC Question



This question was solved and the answer was 30, how it was possible? Need a valid explanation please.






arithmetic recreational-mathematics puzzle







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 2:55












  • $begingroup$
    Why -ve 1, please comment
    $endgroup$
    – Arshad Ali
    Aug 24 '17 at 3:05










  • $begingroup$
    I don't understand the downvotes, either. Added the recreational-mathematics and puzzle tags where the question may be better received.
    $endgroup$
    – dxiv
    Aug 24 '17 at 3:11












  • $begingroup$
    I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 3:12






  • 2




    $begingroup$
    It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
    $endgroup$
    – Bill Dubuque
    Aug 24 '17 at 3:48


















5












$begingroup$


From the days I started to learn Maths, I've have been taught that




Adding Odd times Odd numbers the Answer always would be Odd

e.g 3 + 5 + 1 = 9




OK, but look at this question



UPSC Question



This question was solved and the answer was 30, how it was possible? Need a valid explanation please.






arithmetic recreational-mathematics puzzle







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 2:55












  • $begingroup$
    Why -ve 1, please comment
    $endgroup$
    – Arshad Ali
    Aug 24 '17 at 3:05










  • $begingroup$
    I don't understand the downvotes, either. Added the recreational-mathematics and puzzle tags where the question may be better received.
    $endgroup$
    – dxiv
    Aug 24 '17 at 3:11












  • $begingroup$
    I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 3:12






  • 2




    $begingroup$
    It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
    $endgroup$
    – Bill Dubuque
    Aug 24 '17 at 3:48
















5












5








5


2



$begingroup$


From the days I started to learn Maths, I've have been taught that




Adding Odd times Odd numbers the Answer always would be Odd

e.g 3 + 5 + 1 = 9




OK, but look at this question



UPSC Question



This question was solved and the answer was 30, how it was possible? Need a valid explanation please.






arithmetic recreational-mathematics puzzle







share|cite|improve this question











$endgroup$




From the days I started to learn Maths, I've have been taught that




Adding Odd times Odd numbers the Answer always would be Odd

e.g 3 + 5 + 1 = 9




OK, but look at this question



UPSC Question



This question was solved and the answer was 30, how it was possible? Need a valid explanation please.






arithmetic recreational-mathematics puzzle




arithmetic recreational-mathematics puzzle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 '17 at 3:09









dxiv

57.8k648101




57.8k648101










asked Aug 24 '17 at 2:52









Arshad AliArshad Ali

13514




13514








  • 6




    $begingroup$
    Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 2:55












  • $begingroup$
    Why -ve 1, please comment
    $endgroup$
    – Arshad Ali
    Aug 24 '17 at 3:05










  • $begingroup$
    I don't understand the downvotes, either. Added the recreational-mathematics and puzzle tags where the question may be better received.
    $endgroup$
    – dxiv
    Aug 24 '17 at 3:11












  • $begingroup$
    I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 3:12






  • 2




    $begingroup$
    It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
    $endgroup$
    – Bill Dubuque
    Aug 24 '17 at 3:48
















  • 6




    $begingroup$
    Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 2:55












  • $begingroup$
    Why -ve 1, please comment
    $endgroup$
    – Arshad Ali
    Aug 24 '17 at 3:05










  • $begingroup$
    I don't understand the downvotes, either. Added the recreational-mathematics and puzzle tags where the question may be better received.
    $endgroup$
    – dxiv
    Aug 24 '17 at 3:11












  • $begingroup$
    I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
    $endgroup$
    – Ross Millikan
    Aug 24 '17 at 3:12






  • 2




    $begingroup$
    It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
    $endgroup$
    – Bill Dubuque
    Aug 24 '17 at 3:48










6




6




$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55






$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55














$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05




$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05












$begingroup$
I don't understand the downvotes, either. Added the recreational-mathematics and puzzle tags where the question may be better received.
$endgroup$
– dxiv
Aug 24 '17 at 3:11






$begingroup$
I don't understand the downvotes, either. Added the recreational-mathematics and puzzle tags where the question may be better received.
$endgroup$
– dxiv
Aug 24 '17 at 3:11














$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12




$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12




2




2




$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48






$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48












8 Answers
8






active

oldest

votes


















7












$begingroup$

You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.



You should not believe everything you read in a photo on the internet.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's so kind of you!
    $endgroup$
    – Arshad Ali
    Aug 24 '17 at 3:00



















6












$begingroup$

Clearly impossible in base 10. Can it be done in another number base?



In base 5: $11+11+3=30$



Actually there are many other possibilities!



Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.



It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    May be! this one makes sense a lot.
    $endgroup$
    – Arshad Ali
    Aug 24 '17 at 17:56










  • $begingroup$
    If you do $1+11+13$ you won't even need to use a number twice.
    $endgroup$
    – Jyrki Lahtonen
    Aug 24 '17 at 18:21










  • $begingroup$
    Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
    $endgroup$
    – bluemaster
    Aug 24 '17 at 23:52



















3












$begingroup$

Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    If this is a riddle, I would do : 13,1 + 7,9 + 9






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
      $endgroup$
      – cgiovanardi
      Aug 24 '17 at 18:28








    • 1




      $begingroup$
      Imitation is the sincerest form of flattery, as the saying goes ;-)
      $endgroup$
      – dxiv
      Aug 26 '17 at 0:35





















    1












    $begingroup$


    you can also repeat the numbers




    Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
    decimal separator).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ooh! so Strange...
      $endgroup$
      – Arshad Ali
      Aug 24 '17 at 3:02






    • 2




      $begingroup$
      @ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
      $endgroup$
      – Ross Millikan
      Aug 24 '17 at 3:14










    • $begingroup$
      @RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
      $endgroup$
      – dxiv
      Aug 24 '17 at 3:19








    • 1




      $begingroup$
      Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
      $endgroup$
      – bluemaster
      Aug 25 '17 at 7:50





















    1












    $begingroup$

    What about
    6+9+15=30?



    they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$


      • Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$


      • Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
        $$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.



      Edit:



      Change of base can also be made to work if number repetition were eliminated.
      $$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
        $endgroup$
        – bluemaster
        Aug 25 '17 at 7:29










      • $begingroup$
        @bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
        $endgroup$
        – Eric Towers
        Aug 25 '17 at 13:27



















      -1












      $begingroup$

      It’s $$3_{3} + 3_{3} +3_{3} = 30.$$



      Where $3_{3}$ is read as $3$ base $3$.






      share|cite|improve this answer











      $endgroup$













        Your Answer





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        8 Answers
        8






        active

        oldest

        votes








        8 Answers
        8






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.



        You should not believe everything you read in a photo on the internet.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          That's so kind of you!
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 3:00
















        7












        $begingroup$

        You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.



        You should not believe everything you read in a photo on the internet.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          That's so kind of you!
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 3:00














        7












        7








        7





        $begingroup$

        You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.



        You should not believe everything you read in a photo on the internet.






        share|cite|improve this answer









        $endgroup$



        You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.



        You should not believe everything you read in a photo on the internet.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 24 '17 at 2:55









        60056005

        36.2k751125




        36.2k751125












        • $begingroup$
          That's so kind of you!
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 3:00


















        • $begingroup$
          That's so kind of you!
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 3:00
















        $begingroup$
        That's so kind of you!
        $endgroup$
        – Arshad Ali
        Aug 24 '17 at 3:00




        $begingroup$
        That's so kind of you!
        $endgroup$
        – Arshad Ali
        Aug 24 '17 at 3:00











        6












        $begingroup$

        Clearly impossible in base 10. Can it be done in another number base?



        In base 5: $11+11+3=30$



        Actually there are many other possibilities!



        Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.



        It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          May be! this one makes sense a lot.
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 17:56










        • $begingroup$
          If you do $1+11+13$ you won't even need to use a number twice.
          $endgroup$
          – Jyrki Lahtonen
          Aug 24 '17 at 18:21










        • $begingroup$
          Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
          $endgroup$
          – bluemaster
          Aug 24 '17 at 23:52
















        6












        $begingroup$

        Clearly impossible in base 10. Can it be done in another number base?



        In base 5: $11+11+3=30$



        Actually there are many other possibilities!



        Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.



        It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          May be! this one makes sense a lot.
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 17:56










        • $begingroup$
          If you do $1+11+13$ you won't even need to use a number twice.
          $endgroup$
          – Jyrki Lahtonen
          Aug 24 '17 at 18:21










        • $begingroup$
          Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
          $endgroup$
          – bluemaster
          Aug 24 '17 at 23:52














        6












        6








        6





        $begingroup$

        Clearly impossible in base 10. Can it be done in another number base?



        In base 5: $11+11+3=30$



        Actually there are many other possibilities!



        Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.



        It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.






        share|cite|improve this answer











        $endgroup$



        Clearly impossible in base 10. Can it be done in another number base?



        In base 5: $11+11+3=30$



        Actually there are many other possibilities!



        Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.



        It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 25 '17 at 12:02

























        answered Aug 24 '17 at 13:34









        bluemasterbluemaster

        1,638519




        1,638519












        • $begingroup$
          May be! this one makes sense a lot.
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 17:56










        • $begingroup$
          If you do $1+11+13$ you won't even need to use a number twice.
          $endgroup$
          – Jyrki Lahtonen
          Aug 24 '17 at 18:21










        • $begingroup$
          Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
          $endgroup$
          – bluemaster
          Aug 24 '17 at 23:52


















        • $begingroup$
          May be! this one makes sense a lot.
          $endgroup$
          – Arshad Ali
          Aug 24 '17 at 17:56










        • $begingroup$
          If you do $1+11+13$ you won't even need to use a number twice.
          $endgroup$
          – Jyrki Lahtonen
          Aug 24 '17 at 18:21










        • $begingroup$
          Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
          $endgroup$
          – bluemaster
          Aug 24 '17 at 23:52
















        $begingroup$
        May be! this one makes sense a lot.
        $endgroup$
        – Arshad Ali
        Aug 24 '17 at 17:56




        $begingroup$
        May be! this one makes sense a lot.
        $endgroup$
        – Arshad Ali
        Aug 24 '17 at 17:56












        $begingroup$
        If you do $1+11+13$ you won't even need to use a number twice.
        $endgroup$
        – Jyrki Lahtonen
        Aug 24 '17 at 18:21




        $begingroup$
        If you do $1+11+13$ you won't even need to use a number twice.
        $endgroup$
        – Jyrki Lahtonen
        Aug 24 '17 at 18:21












        $begingroup$
        Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
        $endgroup$
        – bluemaster
        Aug 24 '17 at 23:52




        $begingroup$
        Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
        $endgroup$
        – bluemaster
        Aug 24 '17 at 23:52











        3












        $begingroup$

        Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.






            share|cite|improve this answer









            $endgroup$



            Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 25 '17 at 3:52









            Robert SoupeRobert Soupe

            11k21950




            11k21950























                2












                $begingroup$

                If this is a riddle, I would do : 13,1 + 7,9 + 9






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
                  $endgroup$
                  – cgiovanardi
                  Aug 24 '17 at 18:28








                • 1




                  $begingroup$
                  Imitation is the sincerest form of flattery, as the saying goes ;-)
                  $endgroup$
                  – dxiv
                  Aug 26 '17 at 0:35


















                2












                $begingroup$

                If this is a riddle, I would do : 13,1 + 7,9 + 9






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
                  $endgroup$
                  – cgiovanardi
                  Aug 24 '17 at 18:28








                • 1




                  $begingroup$
                  Imitation is the sincerest form of flattery, as the saying goes ;-)
                  $endgroup$
                  – dxiv
                  Aug 26 '17 at 0:35
















                2












                2








                2





                $begingroup$

                If this is a riddle, I would do : 13,1 + 7,9 + 9






                share|cite|improve this answer









                $endgroup$



                If this is a riddle, I would do : 13,1 + 7,9 + 9







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 24 '17 at 17:30









                cgiovanardicgiovanardi

                749411




                749411












                • $begingroup$
                  You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
                  $endgroup$
                  – cgiovanardi
                  Aug 24 '17 at 18:28








                • 1




                  $begingroup$
                  Imitation is the sincerest form of flattery, as the saying goes ;-)
                  $endgroup$
                  – dxiv
                  Aug 26 '17 at 0:35




















                • $begingroup$
                  You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
                  $endgroup$
                  – cgiovanardi
                  Aug 24 '17 at 18:28








                • 1




                  $begingroup$
                  Imitation is the sincerest form of flattery, as the saying goes ;-)
                  $endgroup$
                  – dxiv
                  Aug 26 '17 at 0:35


















                $begingroup$
                You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
                $endgroup$
                – cgiovanardi
                Aug 24 '17 at 18:28






                $begingroup$
                You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
                $endgroup$
                – cgiovanardi
                Aug 24 '17 at 18:28






                1




                1




                $begingroup$
                Imitation is the sincerest form of flattery, as the saying goes ;-)
                $endgroup$
                – dxiv
                Aug 26 '17 at 0:35






                $begingroup$
                Imitation is the sincerest form of flattery, as the saying goes ;-)
                $endgroup$
                – dxiv
                Aug 26 '17 at 0:35













                1












                $begingroup$


                you can also repeat the numbers




                Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
                decimal separator).






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Ooh! so Strange...
                  $endgroup$
                  – Arshad Ali
                  Aug 24 '17 at 3:02






                • 2




                  $begingroup$
                  @ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
                  $endgroup$
                  – Ross Millikan
                  Aug 24 '17 at 3:14










                • $begingroup$
                  @RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
                  $endgroup$
                  – dxiv
                  Aug 24 '17 at 3:19








                • 1




                  $begingroup$
                  Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
                  $endgroup$
                  – bluemaster
                  Aug 25 '17 at 7:50


















                1












                $begingroup$


                you can also repeat the numbers




                Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
                decimal separator).






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Ooh! so Strange...
                  $endgroup$
                  – Arshad Ali
                  Aug 24 '17 at 3:02






                • 2




                  $begingroup$
                  @ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
                  $endgroup$
                  – Ross Millikan
                  Aug 24 '17 at 3:14










                • $begingroup$
                  @RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
                  $endgroup$
                  – dxiv
                  Aug 24 '17 at 3:19








                • 1




                  $begingroup$
                  Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
                  $endgroup$
                  – bluemaster
                  Aug 25 '17 at 7:50
















                1












                1








                1





                $begingroup$


                you can also repeat the numbers




                Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
                decimal separator).






                share|cite|improve this answer









                $endgroup$




                you can also repeat the numbers




                Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
                decimal separator).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 24 '17 at 3:01









                dxivdxiv

                57.8k648101




                57.8k648101












                • $begingroup$
                  Ooh! so Strange...
                  $endgroup$
                  – Arshad Ali
                  Aug 24 '17 at 3:02






                • 2




                  $begingroup$
                  @ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
                  $endgroup$
                  – Ross Millikan
                  Aug 24 '17 at 3:14










                • $begingroup$
                  @RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
                  $endgroup$
                  – dxiv
                  Aug 24 '17 at 3:19








                • 1




                  $begingroup$
                  Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
                  $endgroup$
                  – bluemaster
                  Aug 25 '17 at 7:50




















                • $begingroup$
                  Ooh! so Strange...
                  $endgroup$
                  – Arshad Ali
                  Aug 24 '17 at 3:02






                • 2




                  $begingroup$
                  @ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
                  $endgroup$
                  – Ross Millikan
                  Aug 24 '17 at 3:14










                • $begingroup$
                  @RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
                  $endgroup$
                  – dxiv
                  Aug 24 '17 at 3:19








                • 1




                  $begingroup$
                  Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
                  $endgroup$
                  – bluemaster
                  Aug 25 '17 at 7:50


















                $begingroup$
                Ooh! so Strange...
                $endgroup$
                – Arshad Ali
                Aug 24 '17 at 3:02




                $begingroup$
                Ooh! so Strange...
                $endgroup$
                – Arshad Ali
                Aug 24 '17 at 3:02




                2




                2




                $begingroup$
                @ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
                $endgroup$
                – Ross Millikan
                Aug 24 '17 at 3:14




                $begingroup$
                @ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
                $endgroup$
                – Ross Millikan
                Aug 24 '17 at 3:14












                $begingroup$
                @RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
                $endgroup$
                – dxiv
                Aug 24 '17 at 3:19






                $begingroup$
                @RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
                $endgroup$
                – dxiv
                Aug 24 '17 at 3:19






                1




                1




                $begingroup$
                Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
                $endgroup$
                – bluemaster
                Aug 25 '17 at 7:50






                $begingroup$
                Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
                $endgroup$
                – bluemaster
                Aug 25 '17 at 7:50













                1












                $begingroup$

                What about
                6+9+15=30?



                they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  What about
                  6+9+15=30?



                  they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    What about
                    6+9+15=30?



                    they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)






                    share|cite|improve this answer









                    $endgroup$



                    What about
                    6+9+15=30?



                    they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 25 '17 at 11:20









                    AndrewAndrew

                    1071212




                    1071212























                        1












                        $begingroup$


                        • Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$


                        • Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
                          $$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.



                        Edit:



                        Change of base can also be made to work if number repetition were eliminated.
                        $$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
                          $endgroup$
                          – bluemaster
                          Aug 25 '17 at 7:29










                        • $begingroup$
                          @bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
                          $endgroup$
                          – Eric Towers
                          Aug 25 '17 at 13:27
















                        1












                        $begingroup$


                        • Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$


                        • Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
                          $$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.



                        Edit:



                        Change of base can also be made to work if number repetition were eliminated.
                        $$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
                          $endgroup$
                          – bluemaster
                          Aug 25 '17 at 7:29










                        • $begingroup$
                          @bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
                          $endgroup$
                          – Eric Towers
                          Aug 25 '17 at 13:27














                        1












                        1








                        1





                        $begingroup$


                        • Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$


                        • Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
                          $$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.



                        Edit:



                        Change of base can also be made to work if number repetition were eliminated.
                        $$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$






                        share|cite|improve this answer











                        $endgroup$




                        • Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$


                        • Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
                          $$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.



                        Edit:



                        Change of base can also be made to work if number repetition were eliminated.
                        $$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Aug 31 '17 at 14:25

























                        answered Aug 25 '17 at 4:18









                        Eric TowersEric Towers

                        32.4k22268




                        32.4k22268












                        • $begingroup$
                          parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
                          $endgroup$
                          – bluemaster
                          Aug 25 '17 at 7:29










                        • $begingroup$
                          @bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
                          $endgroup$
                          – Eric Towers
                          Aug 25 '17 at 13:27


















                        • $begingroup$
                          parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
                          $endgroup$
                          – bluemaster
                          Aug 25 '17 at 7:29










                        • $begingroup$
                          @bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
                          $endgroup$
                          – Eric Towers
                          Aug 25 '17 at 13:27
















                        $begingroup$
                        parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
                        $endgroup$
                        – bluemaster
                        Aug 25 '17 at 7:29




                        $begingroup$
                        parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
                        $endgroup$
                        – bluemaster
                        Aug 25 '17 at 7:29












                        $begingroup$
                        @bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
                        $endgroup$
                        – Eric Towers
                        Aug 25 '17 at 13:27




                        $begingroup$
                        @bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
                        $endgroup$
                        – Eric Towers
                        Aug 25 '17 at 13:27











                        -1












                        $begingroup$

                        It’s $$3_{3} + 3_{3} +3_{3} = 30.$$



                        Where $3_{3}$ is read as $3$ base $3$.






                        share|cite|improve this answer











                        $endgroup$


















                          -1












                          $begingroup$

                          It’s $$3_{3} + 3_{3} +3_{3} = 30.$$



                          Where $3_{3}$ is read as $3$ base $3$.






                          share|cite|improve this answer











                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            It’s $$3_{3} + 3_{3} +3_{3} = 30.$$



                            Where $3_{3}$ is read as $3$ base $3$.






                            share|cite|improve this answer











                            $endgroup$



                            It’s $$3_{3} + 3_{3} +3_{3} = 30.$$



                            Where $3_{3}$ is read as $3$ base $3$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 27 '18 at 1:34









                            dantopa

                            6,46942243




                            6,46942243










                            answered Nov 27 '18 at 0:39









                            Darshan JainDarshan Jain

                            71




                            71






























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