For each positive integer n, let S(n) denote the sum of the digits of n. How many three-digit n's are there such that n == S(n) + S(S(n)) (mod 9)?
+506 The divisibility rule for 9 says that: \(S(n)\equiv n (\text{mod 9})\)
This rule is true every number can be written as \(x_1+10x_2+...10^{n-1}x_n\), where \(x_n\) is the nth digit of the number, and since 10^n is always congruent to 1 mod 9 (because 10^n-1 is 99...9, which is divisible by 9), the expression reduces to \(x_1+1x_2+...1x_n\), which is simply the sum of the digits.
Anyway, back to the question, the original expression reduces to
\(n \equiv n+n (\text{mod 9})\\n\equiv 0 (\text{mod 9})\)
Now, we just need to find the number of 3 digit multiples of 9. Can you finish it?
+506 The divisibility rule for 9 says that: \(S(n)\equiv n (\text{mod 9})\)
This rule is true every number can be written as \(x_1+10x_2+...10^{n-1}x_n\), where \(x_n\) is the nth digit of the number, and since 10^n is always congruent to 1 mod 9 (because 10^n-1 is 99...9, which is divisible by 9), the expression reduces to \(x_1+1x_2+...1x_n\), which is simply the sum of the digits.
Anyway, back to the question, the original expression reduces to
\(n \equiv n+n (\text{mod 9})\\n\equiv 0 (\text{mod 9})\)
Now, we just need to find the number of 3 digit multiples of 9. Can you finish it?
