2007 AMC 10A Problem 25

Bound the search. Since $n < 2007$, $n$ has at most four digits with the leading one at most $2$; the largest digit sum available is $S(1999) = 28$, and then $S(S(n)) \le S(28) = 10$ (in fact at most $10$ for any $S(n) \le 28$). So
$$n = 2007 - S(n) - S(S(n)) \ge 2007 - 28 - 10 = 1969.$$

Filter by $9$. A number is congruent to its digit sum modulo $9$, so $n \equiv S(n) \equiv S(S(n)) \pmod 9$. The equation gives $3n \equiv 2007 \equiv 0 \pmod 9$, hence $n \equiv 0 \pmod 3$.

That leaves the multiples of $3$ from $1971$ to $2007$, thirteen candidates. Compute $n + S(n) + S(S(n))$ for each:

Exactly four values work: $n = 1977, 1980, 1983, 2001$.

The answer is $\boxed{\textbf{(D)}\ 4}$.

Digit-sum equations are tamed by two tools: a size bound (digit sums are tiny compared with the number) that restricts $n$ to a short interval, and the mod-$9$ invariant $n \equiv S(n)$ that thins the interval further. What remains is a checklist small enough to finish by hand. Whenever $S(n)$ appears, think "mod 9" and "at most $9 \times$ (number of digits)."

• Testing every $n$ from $1969$ to $2007$ without the mod-$9$ filter, or dropping the $S(S(n))$ term when estimating the bound and starting the search too high.
• Concluding $n \equiv 0 \pmod 9$ from $3n \equiv 0 \pmod 9$, which would wrongly discard $1977, 1983$ and $2001$ and leave only $1980$.