Sum of the digits | AMC-10A, 2007 | Problem 25

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

Let \(P(n)=(n+s(n)+s(s(n))=2007)\) tnen obviously \(n<2007\)

For \(n\)=\(1999\).the sum becoms \(28+10)=38\)

so we may say that the minimum bound is \(1969\)

Now we want to break it in 3 parts .....

Case 1:\(n \geq 2000\),

Case 2:\(n \leq 2000\) (\(n = 19xy,x+y<10\)

Case 3:\(n \leq 2000\) (\(n = 19xy,x+y \geq 10\)

Can you now finish the problem ..........

Case 1:\(n \geq 2000\),

Then \(P(n)=(n+s(n)+s(s(n))=2007)\) gives \(n=2001\)

Case 2:\(n \leq 2000\) (\(n = 19xy,x+y<10\)

Then \(P(n)=(n+s(n)+s(s(n))=2007)\) gives \(4x+y=32\) which satisfying the constraints \(x = 8\), \(y = 0\).

Case 3:\(n \leq 2000) ((n = 19xy,x+y \geq 10\) gives \(4x+y=35\) which satisfying the constraints \(x = 7\), \(y = 7\) and \(x = 8\), \(y = 3\).

can you finish the problem........

Therefore The solutions are thus \(1977, 1980, 1983, 2001\)

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