Q. Let n be the largest integer for which 14n has exactly 100 digits. Counting from right to left, find the 68th digit of n

• We simply know that 999999....(100 times)is the greatest integer of 100-digits.Now let's assume A=999999....(100 times); If we start dividing 'A' by 14 ,after a while we easily observe that 9999999= 14*714285 + 9;Now carrying this remainder '9' along with next six 9's of 'A' and dividing it by 14 will again produce the same result,i.e.,
  9999999999999(i.e. thirteen 9's)=14*(714285714285) + 9; And also for the next six 9's ,i.e.,when we'll divide 1st nineteen 9's of 'A',we can predict the result at ease that we'll obtain quotient as (714285714285714285) & remainder will be 9..At this point we can easily Predict the quotient & remainder when {9999999999999999999......(97 times 9)} will be divided by 14 and the quotient & remainder will be {(714285)(714285)(714285)....16 times} and 9 respectively and this remainder is the 97th 9 of 'A'.And we still have 9's at the 98th,99th &100th positions.We clearly observe that 9999=14*(811)+5; i.e,5 is the remainder when we divide 'A' by 14.Therefore if 'n' be the largest integer with 14n containing 100 digits ,then 14n = A-5,which implies
  n=(999999999999....{99 times 9}4) and the quotient is
  [{(714285)(714285)(714285)....16 times}811]..Now from right side we can easily count the 68th position and accurately find the 68th digit..since each of [(714285)]-this block contains 6 digits,therefore counting from right to left side ,the [11*6 +3]=69th digit will be 7,therefore ,the 68th digit will definitely be '1'.
  So,Counting from right to left,the 68th digit will be '1'.
• 11 years agoHelpfull: Yes(8) No(0)
• we have to determine the largest integer n for which 14n < 10^100, since 14n
  is an integer.
  
  largest integer n for which n < (10^100)/14 = (5/7)*10^99 = 0.714285714285....*10^99 where term 714285 is repeating after decimal point.
  
  considering this , we find that 68th digit from the right is 1.
• 12 years agoHelpfull: Yes(1) No(6)
• @ DIPIN sir as 14n is an integer so i think it can't be exactly 0.714285714285....*10^99 because here 0.714285... and it will repeat itself infinite times.. so you can't say surely that 68th digit will be 1.
  
  now i am trying to correct it .. please see this correction in the solution
  as
  14n is an integer of exactly 100 digits and here n is largest integer so in other way we can say 14n is largest perfect integer of 100 digits and
  we know that largest integer of 100 digits will be
  9999999999.........99 (100 times)
  which is equal to = 9(1111....1111111) (100 times 1)
  = 9 (10^100 +10^99 +10^98 +......+10^2 +10 +1)
  =9*(10^101 -1)/(10-1)
  =(10^101 -1)
  so in this way largest integer of 100 digits will be equal to (10^101 -1) number.
  so 14 n < (10^101 -1)
  and hence
  n< (10^101 -1)/14
  by property of inequalities
  n
• 11 years agoHelpfull: Yes(1) No(2)
• I didn.t understand Dipin,
  Could u pl clarify
• 12 years agoHelpfull: Yes(0) No(1)
• my answer is not showing completely i don't know why ?
• 11 years agoHelpfull: Yes(0) No(0)