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

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 $n+s(n)+s(s(n))=2007$

algebra

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multiplication

Answer: $4$

AMC-10A (2007) Problem 25

Pre College Mathematics

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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