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Thread: Count the Digits

  1. #1
    Junior Member
    Join Date
    Jul 2007

    Question Count the Digits

    Can u find a number which added to itself one or several times will give a total having the same digits as that number but differently arranged and after the sixth edition will give a total of all nines?

  2. #2

    Re: Count the Digits

    142857 is the number

    142857 which added for six times will give only the six digit answer and added for the seventh time will give the answer 999999 - all nines.

    i. 142857+142857=285714
    ii. 142857+142857+142857=428571
    iii. 142857+142857+142857+142857=571428
    iv. 142857+142857+142857+142857+142857=714285
    vi. 142857+142857+142857+142857+142857+142857+142857=999999

  3. #3
    Junior Member
    Join Date
    Jul 2007

    Re: Count the Digits

    its correct

  4. #4
    Junior Member
    Join Date
    Jul 2008

    Re: Count the Digits

    hii guys here is a doubt for me you are giving answers but can we remember as such
    142857 so give us method to find the number tooo...............

    heres my method for this

    first we must get 9 in the last digit after six additions of same number meaning the last digit of the number multiplied by 7 must give 9 in the last position of resultant so it means only 7 is possible as last digit in the we place 7 in last digit position...


    now,we get carry 4 (7*7=49) to next digit so we need only 5 in next digit.......
    to get 5 only 5 is possible(since 7*5=35)


    now,we get carry 3(7*5=35) to next digit so we need only 6 in next digit...
    to get 6 only 8 is possible(since 7*8=56)


    next,similarly carry:5 rdigit:4 nextdigit:2(7*2=14)


    next,carry:1 rdigit:8 nextdigit:4(7*4=28)


    next carry:2 rdigit:7 nextdigit:1(7*1=7)


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