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in Olympiad by (76.3k points)

For any natural number n, let S(n) denote the sum of the digits of n. Find the number of all 3-digit numbers n such that S(S(n)) = 2. 

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Let n = abc 

where n = a × 100 + b × 10 + c a three digit number 

a + b + c = S(n); Here note that S(n) ≤ 27 

Since S(S(n)) = 2, 

It means sum of digits of S(n) is 2 

Now, S(n) can be, S(n) = 2, 11, 20 only

Now Case-1 

 a + b + c = 2; possible cases are {0, 1, 1} gives 2 number and {2,0, 0} gives 1 number (ex : 200) 

Total number in case (1) are 3 

Total in case (3) = (3 × 4 + 6 × 4) = 36 ways 

Total numbers = 3 + 61 + 36 = 100 

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