Try this beautiful problem from Algebra based on Sum of the digits.

## 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 \(n+s(n)+s(s(n))=2007\)

- \(1\)
- \(2\)
- \(3\)
- \(4\)

**Key Concepts**

algebra

function

multiplication

## Check the Answer

But try the problem first…

Answer: \(4\)

AMC-10A (2007) Problem 25

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

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……..

Final Step

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

## Other useful links

- https://www.cheenta.com/geometry-based-on-triangle-prmo-2018-problem-63/
- https://www.youtube.com/watch?v=7AlfBAPWEMg

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