Prove that number $underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}$ is product of two consecutive...











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Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers



$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.



Is this good path or not?






discrete-mathematics







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  • 2




    You need a backslash before underbrace to get what you want.
    – Ross Millikan
    Nov 16 at 15:43






  • 2




    I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
    – lulu
    Nov 16 at 15:45















up vote
1
down vote

favorite












Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers



$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.



Is this good path or not?






discrete-mathematics







share|cite|improve this question




















  • 2




    You need a backslash before underbrace to get what you want.
    – Ross Millikan
    Nov 16 at 15:43






  • 2




    I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
    – lulu
    Nov 16 at 15:45













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers



$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.



Is this good path or not?






discrete-mathematics







share|cite|improve this question















Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers



$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.



Is this good path or not?






discrete-mathematics




discrete-mathematics






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edited Nov 16 at 16:15









Yadati Kiran

1,237417




1,237417










asked Nov 16 at 15:41









Marko Škorić

69810




69810








  • 2




    You need a backslash before underbrace to get what you want.
    – Ross Millikan
    Nov 16 at 15:43






  • 2




    I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
    – lulu
    Nov 16 at 15:45














  • 2




    You need a backslash before underbrace to get what you want.
    – Ross Millikan
    Nov 16 at 15:43






  • 2




    I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
    – lulu
    Nov 16 at 15:45








2




2




You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43




You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43




2




2




I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45




I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45










3 Answers
3






active

oldest

votes

















up vote
1
down vote



accepted










Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$



$$begin{align}
&dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
= &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
= &(b+1), b
end{align}$$






share|cite|improve this answer




























    up vote
    2
    down vote













    You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.






    share|cite|improve this answer



















    • 2




      $111222=333times334$
      – Yadati Kiran
      Nov 16 at 15:47








    • 2




      @YadatiKiran: that is correct, but OP has not demonstrated that yet.
      – Ross Millikan
      Nov 16 at 15:48


















    up vote
    0
    down vote













    Insert $1=frac{1}{3}*3$



    $$
    (10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
    x= frac{10^{100}+2}{3}\
    y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
    $$






    share|cite|improve this answer





















      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$



      $$begin{align}
      &dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
      = &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
      = &(b+1), b
      end{align}$$






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$



        $$begin{align}
        &dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
        = &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
        = &(b+1), b
        end{align}$$






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$



          $$begin{align}
          &dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
          = &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
          = &(b+1), b
          end{align}$$






          share|cite|improve this answer












          Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$



          $$begin{align}
          &dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
          = &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
          = &(b+1), b
          end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 16 at 16:25









          Bill Dubuque

          207k29189624




          207k29189624






















              up vote
              2
              down vote













              You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.






              share|cite|improve this answer



















              • 2




                $111222=333times334$
                – Yadati Kiran
                Nov 16 at 15:47








              • 2




                @YadatiKiran: that is correct, but OP has not demonstrated that yet.
                – Ross Millikan
                Nov 16 at 15:48















              up vote
              2
              down vote













              You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.






              share|cite|improve this answer



















              • 2




                $111222=333times334$
                – Yadati Kiran
                Nov 16 at 15:47








              • 2




                @YadatiKiran: that is correct, but OP has not demonstrated that yet.
                – Ross Millikan
                Nov 16 at 15:48













              up vote
              2
              down vote










              up vote
              2
              down vote









              You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.






              share|cite|improve this answer














              You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 16 at 15:48

























              answered Nov 16 at 15:46









              Ross Millikan

              288k23195365




              288k23195365








              • 2




                $111222=333times334$
                – Yadati Kiran
                Nov 16 at 15:47








              • 2




                @YadatiKiran: that is correct, but OP has not demonstrated that yet.
                – Ross Millikan
                Nov 16 at 15:48














              • 2




                $111222=333times334$
                – Yadati Kiran
                Nov 16 at 15:47








              • 2




                @YadatiKiran: that is correct, but OP has not demonstrated that yet.
                – Ross Millikan
                Nov 16 at 15:48








              2




              2




              $111222=333times334$
              – Yadati Kiran
              Nov 16 at 15:47






              $111222=333times334$
              – Yadati Kiran
              Nov 16 at 15:47






              2




              2




              @YadatiKiran: that is correct, but OP has not demonstrated that yet.
              – Ross Millikan
              Nov 16 at 15:48




              @YadatiKiran: that is correct, but OP has not demonstrated that yet.
              – Ross Millikan
              Nov 16 at 15:48










              up vote
              0
              down vote













              Insert $1=frac{1}{3}*3$



              $$
              (10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
              x= frac{10^{100}+2}{3}\
              y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
              $$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Insert $1=frac{1}{3}*3$



                $$
                (10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
                x= frac{10^{100}+2}{3}\
                y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
                $$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Insert $1=frac{1}{3}*3$



                  $$
                  (10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
                  x= frac{10^{100}+2}{3}\
                  y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
                  $$






                  share|cite|improve this answer












                  Insert $1=frac{1}{3}*3$



                  $$
                  (10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
                  x= frac{10^{100}+2}{3}\
                  y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 16 at 15:51









                  AHusain

                  2,7602816




                  2,7602816






























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