Problem:
For each positive integer n, let S(n) denote the sum of the digits of n. For how many values of n is n+S(n)+S(S(n))=2007?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
If n≤2007, then S(n)≤S(1999)=28. If n≤28, then S(n)≤S(28)=10. Therefore if n satisfies the required condition it must also satisfy
n≥2007−28−10=1969
In addition, n,S(n), and S(S(n)) all leave the same remainder when divided by 9. Because 2007 is a multiple of 9, it follows that n,S(n), and S(S(n)) must all be multiples of 3. The required condition is satisfied by (D)4​ multiples of 3 between 1969 and 2007, namely 1977,1980,1983, and 2001.
Note: There appear to be many cases to check, that is, all the multiples of 3 between 1969 and 2007. However, for 1987≤n≤1999, we have n+S(n)≥ 1990+19=2009, so these numbers are eliminated. Thus we need only check 1971,1974,1977,1980,1983,1986,2001, and 2004.
The problems on this page are the property of the MAA's American Mathematics Competitions