+0

# number theory

0
42
1

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)?

Jun 25, 2021

#1
+420
+1

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?

Jun 25, 2021

#1
+420
+1

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?

textot Jun 25, 2021