Find the 100th place digit of a number forming from digits 1 to 9 without repeatation and number passes the

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.
8 years agoHelpfull: Yes(15) No(16)
Please correct the question "The number should be greatest from all such numbers satisfying that conditions"
8 years agoHelpfull: Yes(9) No(4)
Largest no is 987652413
So digit will be 4
8 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
7 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
8 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.
8 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
8 years agoHelpfull: Yes(0) No(10)

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.