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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)?

 Dec 3, 2021
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Quote from Google

"If a number is being divided by 9, add each of the digits to each other until you are left with one number (e.g., 1164 becomes 12 which in turn becomes 3), which is the remainder."

 

n (mod9) = S(n) (mod 9) = S(S(n)) (mod 9)     for all integer n

 

so

n = S(n) + S(S(n)) (mod  9)

becomes

n (mod9) = n (mod9) +n (mod9) 

2n=n (mod9)

2n=n+9k 

 

n=9k

n must be a muliple of 9

How many multiples of 9 are there between 100 and 999?

 

111-10  =  101  

 

If anyone thinks I have made a logic error please let me know, with your reasons of course.

 Dec 3, 2021

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