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steve jobs

17/03/2017 at 14:19
Answers
4
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111...111 (2002 1s) 555...555 (2002 5s) is the product of two consecutive odd numbers.

Find the sum of these two odd numbers.


rational numbers


    List of answers
  • ...
    Dung Trần Thùy 18/03/2017 at 18:54

    Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;

    333 . 335 = 111555, ... It's the first step to solve ur problem.

    Now, we have to prove that expression equal 333...3 . 333....5.

    We have :

    \(111...11111111111555...555555555\)

    ( 2002 1s)                      (2002 5s)

    =111.....11000....0 + 555.......5 

    ( 2002 1s) (2002 0s) (2002 5s)

    = 1111.....111 . ( 10000...000 + 5 )

        ( 2002 1s)          ( 2002 0s)

    = 111....111 .  10000...00005

        ( 2002 1s)       (2001 0s )

    = 1111...1111 . ( 3 . 333...33335 )

      ( 2002 1s)               (2001 3s )

    = 333......3333 . 333333...3335

           ( 2002 3s)     ( 2001 3s )

    The sum of these 2 numbers is 6666......68

                                                      (2002 6s)

    Sorry if my English is bad :>

    steve jobs selected this answer.
  • ...
    Lê Anh Tú 25/03/2017 at 22:18

    Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;

    333 . 335 = 111555, ... It's the first step to solve ur problem.

    Now, we have to prove that expression equal 333...3 . 333....5.

    We have :

    111...11111111111555...555555555111...11111111111555...555555555

    ( 2002 1s)                      (2002 5s)

    =111.....11000....0 + 555.......5 

    ( 2002 1s) (2002 0s) (2002 5s)

    = 1111.....111 . ( 10000...000 + 5 )

        ( 2002 1s)          ( 2002 0s)

    = 111....111 .  10000...00005

        ( 2002 1s)       (2001 0s )

    = 1111...1111 . ( 3 . 333...33335 )

      ( 2002 1s)               (2001 3s )

    = 333......3333 . 333333...3335

           ( 2002 3s)     ( 2001 3s )

    The sum of these 2 numbers is 6666......68

                                                      (2002 6s)

  • ...
    FA KAKALOTS 03/02/2018 at 12:46

    Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;

    333 . 335 = 111555, ... It's the first step to solve ur problem.

    Now, we have to prove that expression equal 333...3 . 333....5.

    We have :

    111...11111111111555...555555555

    ( 2002 1s)                      (2002 5s)

    =111.....11000....0 + 555.......5 

    ( 2002 1s) (2002 0s) (2002 5s)

    = 1111.....111 . ( 10000...000 + 5 )

        ( 2002 1s)          ( 2002 0s)

    = 111....111 .  10000...00005

        ( 2002 1s)       (2001 0s )

    = 1111...1111 . ( 3 . 333...33335 )

      ( 2002 1s)               (2001 3s )

    = 333......3333 . 333333...3335

           ( 2002 3s)     ( 2001 3s )

    The sum of these 2 numbers is 6666......68

                                                      (2002 6s)

    Sorry if my English is bad :>

  • ...
    NGUYỄN SANH KIÊN 18/03/2017 at 06:32

    236/3


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