Find the 100th place digit of a number forming from digits 1 to 9 without repeatation and number passes the divisibility test of 11 means difference of sum of alternative digits is 0 or divisible by 11.

• Largest no is 987652413
  So digit will be 4
• 7 years agoHelpfull: Yes(27) No(5)
• The 100th place digit would be 4.
  As the largest no. which is divisible by 11 having digits 1 to 9 is 978653412 ,
  the difference of the sum of the digits at even place and odd place is 11, which is a multiple of 11.
• 7 years agoHelpfull: Yes(15) No(16)
• Please correct the question "The number should be greatest from all such numbers satisfying that conditions"
• 7 years agoHelpfull: Yes(9) No(4)
• Largest no is 987652413
  So digit will be 4
• 7 years agoHelpfull: Yes(4) No(0)
• largest no. formed will be 987652413,not 978653412 just because solution on elitmus.tutorial it is being mentioned as such,they have solved it wrong
• 6 years agoHelpfull: Yes(2) No(0)
• Can anyone please explain me how u ppl find the largest no(987652413) that's divisible by 11?
• 6 years agoHelpfull: Yes(2) No(2)
• greatest number is 979
• 7 years agoHelpfull: Yes(0) No(11)
• The 100th place digit would be 4.
  As the largest no. which is divisible by 11 having digits 1 to 9 is 978653412 ,
  the difference of the sum of the digits at even place and odd place is 11, which is a multiple of 11.
• 7 years agoHelpfull: Yes(0) No(2)
• ans will be 1
  becoz
  if we use cyclic method then we get 7^4 which results in 1 at unit place
• 7 years agoHelpfull: Yes(0) No(10)