Problem:
In their base 10 representations, the integer a consists of a sequence of 1985 eights and the integer b consists of a sequence of 1985 fives. What is the sum of the digits of the base 10 representation of the integer 9ab?
Answer Choices:
A. 15880
B. 17856
C. 17865
D. 17874
E. 19851
Solution:
First note that since 9ab ends in a zero, the sum of its digits is the same as the sum of the digits in N=9ab/10. Second note that for any integer M represented as a string of k copies of the digit d, M=9d(999…99)=9d(10k−1). Thus
N=109(98(101985−1))(95(101985−1))=94(102⋅1985−2⋅101985+1)=94((102⋅1985−1)−2(101985−1))=94(102⋅1985−1)−98(101985−1)=P−Q
where P is the number consisting of a sequence of 2⋅1985 fours and Q consists of 1985 eights. Thus N consists of 1984 fours followed by 1 three followed by 1984 fives followed by 1 six. The sum of the digits in N is
1984⋅(4+5)+(6+3)=1985⋅9=17865
Alternatively, one can replace k=1985 by k=1,2,3 and look for a pattern. When k=1,9ab=360 and the sum of the digits is 9. When k=2,9ab=9.88⋅55=43560 and the digit sum is 18. When k=3,9ab= 9⋅888⋅555=4435560 and the digit sum is 27. It now seems clear (and can be proved by induction) that in general the digit sum is 9k.