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The final single digit number will have same remainder on division with 9 as the original number we start with. This means that we need to find number of positive integers which leave a remainder of 5 on division with 9. This is an AP with 1st term as 5 and last term as 995 and common difference 9..
So 9 9 5 = 5 + ( n − 1 ) 9 which gives n = 1 1 1