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)≡n(mod 9)
This rule is true every number can be written as x1+10x2+...10n−1xn, where xn 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 x1+1x2+...1xn, which is simply the sum of the digits.
Anyway, back to the question, the original expression reduces to
n≡n+n(mod 9)n≡0(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)≡n(mod 9)
This rule is true every number can be written as x1+10x2+...10n−1xn, where xn 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 x1+1x2+...1xn, which is simply the sum of the digits.
Anyway, back to the question, the original expression reduces to
n≡n+n(mod 9)n≡0(mod 9)
Now, we just need to find the number of 3 digit multiples of 9. Can you finish it?
