2007 AMC 12A Problem 22

Below is the professionally curated solution for Problem 22 of the 2007 AMC 12A, 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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Concepts:digits modular arithmetic bounding to limit cases

Difficulty rating: 1910

22.

For each positive integer $n,$ let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n+S(n)+S(S(n))=2007?$

$1$

$2$

$3$

$4$

$5$

Solution:

For $n\le 2007,$ $S(n)\le S(1999)=28,$ and then $S(S(n))\le S(28)=10.$ So any solution has $n\ge 2007-28-10=1969.$

Also $n,$ $S(n),$ and $S(S(n))$ are congruent modulo $9,$ and $2007$ is a multiple of $9,$ so all three must be multiples of $3.$

Checking the multiples of $3$ between $1969$ and $2007$ (many are eliminated because $n+S(n)$ already exceeds $2007$ ) leaves $1977,1980,1983,$ and $2001.$ That is $4$ values.

Thus, the correct answer is D.

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