If # is an operation which results in adding the digits of integer until a single digit is left, what is the probability that a number picked up in first 90 positive integers will have the result of # as an odd digit ?
( example 99 = 9+9 = 18=1+8=9, so #=9)
Select:
A) 4/10
B) 4/9
C) 1/2
D) 6/10
E) 5/9
First of all let us find out what is the maximum sum we will get by adding digits of numbers from 1 to 90.   That will be 17 (8+9+17) Why did I calculate this? I will come to that in a moment.   So to get an odd digit by carrying out the operation #, the sum of the digits of the integer should be 1, 3, 5, 7, 9, 10, 12, 14, and 16 (Why not 18? Because we know the maximum sum can be 17)   Number of integers, whose digits add to 1: 1 and 10 = 2   Number of integers, whose digits add to 3: 3, 12, 21, 30 = 4   Number of integers, whose digits add to 5: 5, 14, 23, 32, 41, 50 = 6   Number of integers, whose digits add to 7: 7, 16, 25, 34, 43, 52, 61, 70 = 8   Number of integers, whose digits add to 9: 10 multiples of 9 up to 90 = 10   Number of integers, whose digits add to 10: 19, 28, 37, 46, 55, 64, 73, 82 = 8   Number of integers, whose digits add to 12: 39, 48, 57, 66, 75, 84 = 6   Number of integers, whose digits add to 14: 59, 68, 77, 86 = 4   Number of integers, whose digits add to 16: 79, 88 = 2   So in total there are  2+4+6+8+10+8+6+4+2= 
<gadget id="calculator"> 2+4+6+8+10+8+6+4+2</gadget>
<output>50</output>
50 such numbers   Probability= 50/90 = 5/9   Correct answer 5/9
<result>E</result>
If # is an operation which results in adding the digits of integer until a single digit is left, what is the probability that a number picked up in first 90 positive integers will have the result of # as an odd digit ?
( example 99 = 9+9 = 18=1+8=9, so #=9)