2007 AMC 12A Problem 22

Attempt Problem 22 of the 2007 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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22.

For each positive integer n,n, let S(n)S(n) denote the sum of the digits of n.n. For how many values of nn is n+S(n)+S(S(n))=2007?n+S(n)+S(S(n))=2007?
11
22
33
44
55
Answer: D
Concepts:digitsmodular arithmeticbounding to limit cases
Small hint:
All three of n,n, S(n),S(n), and S(S(n))S(S(n)) leave the same remainder modulo 99
Big hint:
Since 20072007 is a multiple of 9,9, each is a multiple of 33; also n≥2007−28−10n\ge 2007-28-10
Solution:
For n≤2007,n\le 2007, S(n)≤S(1999)=28,S(n)\le S(1999)=28, and then S(S(n))≤S(28)=10.S(S(n))\le S(28)=10. So any solution has n≥2007−28−10=1969.n\ge 2007-28-10=1969.
Also n,n, S(n),S(n), and S(S(n))S(S(n)) are congruent modulo 9,9, and 20072007 is a multiple of 9,9, so all three must be multiples of 3.3.
Checking the multiples of 33 between 19691969 and 20072007 (many are eliminated because n+S(n)n+S(n) already exceeds 20072007) leaves 1977,1980,1983,1977,1980,1983, and 2001.2001. That is 44 values.
Thus, the correct answer is D.
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