Wednesday, September 21, 2011

Problem of the Week 21 Sept 2011

The leftmost digit of an integer of length 2000 digits is 3. In this integer, any two consecutive digits must be divisible by 17 or 23. The 2000th digit may be either a or b. What is the value of a+b?



7 comments:

  1. can someone give a hint

    ReplyDelete
  2. last 5 digits should be 34685 or 34692
    hence a+b is 7

    ReplyDelete
  3. 3468517_ ? This can't be a solution right as you can't find a digit next to 7 to satisfy the given conditions?

    ReplyDelete
  4. The answer is 7.. Just do a recursive loop

    ReplyDelete
  5. See here is the full solution

    first three digits are 346

    fourth can be 8 or 9
    let it be 9
    then fifth is 2, so the number is 34692
    sixth has to be 3 then the solution repeats

    let the fourth digit be 8
    then fifth is 5 sixth is 1 seventh is 7 but no digit after that..

    Hence the recursive sequence of 5 digit appears..

    but the last 5 can be 34685 or 34692.. Hence the sum is 2+5= 7

    ReplyDelete