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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

Best Answer 

 #1
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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
avatar+420 
+1
Best Answer

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

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