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2007 AMC 10A Problem 25

Problem 25 of 25HarderNumber TheoryProblem-Solving Techniques

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?

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Solution

If n≤2007,n \le 2007, then S(n)≤28S(n) \le 28 and S(S(n))≤10,S(S(n)) \le 10, so n≥2007−28−10=1969.n \ge 2007 - 28 - 10 = 1969. Since n,n, S(n),S(n), and S(S(n))S(S(n)) all leave the same remainder modulo 99 and 20072007 is a multiple of 9,9, each must be a multiple of 3.3. Checking the multiples of 33 between 19691969 and 2007,2007, the condition holds for 1977,1977, 1980,1980, 1983,1983, and 2001.2001. So there are 44 values of n.n. Thus, the correct answer is D.
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Tagged: digits · modular arithmetic · bounding to limit cases

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