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