2n + S(n) = 1998.
C) 17 not possible. then sum will be odd.
E) 20. n=989. S(n)!=20
D) 22. n=988 not
A) 24. 987. sum=24
IMO A

The detailed solution to this question would be like this:
lets say the no. is xyz (it cant be a 2 digit no. otherwise the summation wont add upto 1998).
as per the equation 2n+S(n)=1998=> 2(100x+10y+z) + x+y+z=1998
=>201x+21y+3z=1998
=>67x+7y+z=666
To solve above equation:
Lets assume the max value of y and z (which is 9) and it would give us 67x=594
and thus we get the least integral value of x=9.
7y+z=63.
the solution set for (y,z) would be (9,0) and (8,7).
First set gives us the value of n as 990 which doesnt satisfy the eqn above.
Second set gives us the value 987 which satisfy the eqn above.
S(n) is 24.

Cheers
Ami/-

This is one in which I would look at the answers to help if 2n + S(n) = 1998 that means that
1998 - S(n) = 2n, which means that 1998 - S(n) will be even. Then we can eliminate the odd . answers becuase if we subtract an odd number from 1998 we will get an odd number.

It is acutally pretty easy to look at the answers:

24) 1998 - 24 = 1974 - divide that by 2 to get 987 add the digits of 987 and you get 24 - thus this answer works.

22) 1998 - 22 = 1978 - divide by 2 to get 988 - the sum of the digits of 988 is 25 (this doesn't work)

20) 1998 - 20 = 1978 divide by 2 to get 989 - sum of digits is 26 - doesn't match.