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2001 digits (Posted on 2002-04-29) Difficulty: 2 of 5
A string of 2001 digits begins with a "6". Any number formed by two consecutive digits is divisible by either 17 or 23.

What is the last digit in this sequence?

What if the sequence had 2002 digits?

  Submitted by levik    
Rating: 4.3333 (6 votes)
Solution: (Hide)
First write down all the two digit multiples of both numbers:
  • 17: 17, 34, 51, 68, 85
  • 23: 23, 46, 69, 92

    Now, starting with a 6, there are two paths we can take (since 2 of the above numbers start with 6):

  • A: 6 8 5 1 7 - here we come to a dead end, since no two digit multiples of either 17 or 23 start with a 7
  • B: 6 9 2 3 4 6 - in this case, the 6th digit in the sequence is a 6 again, so we come back to the original problem.

    It is obvious that we can repeat pattern B as many times as we need, and come back to a 6 every 5 digits. So a digit number n where n mod 5 = 1 will always be a 6, and since 2001 mod 5 = 1 the 2001st digit is a 6.

    The 2002nd digit can be a 9, but can also be an 8, since we can branch into the dead end pattern A here because we have no need to continue the sequence further.

  • Comments: ( You must be logged in to post comments.)
      Subject Author Date
    SolutionPuzzle SolutionK Sengupta2008-07-21 04:27:44
    answerK Sengupta2007-12-23 11:58:55
    vivek ..this is for u ............ravi2004-09-30 12:02:25
    SolutionsolutionFatBoy2003-08-12 09:40:50
    Some ThoughtsOkGamer2003-04-14 11:33:05
    clarificationRick2003-02-07 05:26:23
    missing pointvivek2002-05-01 06:57:27
    not exactlylevik2002-05-01 05:53:35
    solutionDave Turner2002-04-30 17:15:16
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