Test – 2nd June-Level 1B

Home Forums Math Olympiad, I.S.I., C.M.I. Entrance RMO Test – 2nd June-Level 1B

Viewing 2 posts - 1 through 2 (of 2 total)
  • Author
  • #29151
    Aruna Ramachandran

    A two digit number  formed by any 2 adjacent digits of a 2017 digit umber is divisible by 17 or 23. if the last digit of the 2017 digit number is 1.Find the first digit?

    Reverse the digits 1746 and we get 6741, the new number is larger than the original number by 4725. How many four digit number can satisfy such condition?



    solution of problem 1 )

    working backward 

    see that 17 gives 17, 34,51,68,85 as its two digit multiple

    and 23 gives 23,56,69,92 as its two digit multiple

    now last digit is 1 so the second last digit must be 5

    now if it 5 then the third last digit is 8

    then the 4th last digit must be 6

    now if it is 6 then 5th last digit is 5 again

    so now we get back to 5 again

    so there will be a pattern like this ……6856856851

    using this pattern we can conclude that the 1st digit must be 6 ( why? think with division by 3)

    so , we are done

    • This reply was modified 11 months, 3 weeks ago by Aritra.
Viewing 2 posts - 1 through 2 (of 2 total)
  • You must be logged in to reply to this topic.