Question

A certain number on being divided successively by 9,11 and 13 leaves remainder 8,9 and 8 respectively. what are the remainders when the same

Answer 1

the question is incomplete

Answer 2

Number leaves same reminder when decided by 13&9. so number must be multiple of L.C.M. of that two number plus 8.  »  13*9*n 
--Number also leaves reminder 9 When devided by 11
»  11*k + 9

Comparing both results we get,
117*n + 8 = 11*k + 9
117*n - 1 = 11*k
Above equation can be solved by taking different  values of 'n' Starting from n=1 so that  " 117*n-1 " will be multiple of 11.
» n= 8  gives Multiple of 11 which is 935
Now, 117*n  - 1 = 935
»  117*n + 8 =944 is the smallest number satisfying conditions given in question.
  » Reversing the number will give 449
» Reminders when 449 devided by 11, 9 and 13 will be 9,8 and 7 respectively.