For each positive integer n, let S(n) denote the sum of the digits of

Question

For each positive integer n, let S(n) denote the sum of the digits of n. For how many values of n is n + S(n) + S(S(n)) = 2007?

(A) 1 
(B) 2 
(C) 3 
(D) 4 
(E) 5

Chethan92 · Answer

Got 4 number, which satisfies. 1977+24+6 = 2007
1980+18+9 = 2007
1983+21+3 = 2007
2001+3+3 = 2007
I chose a number, which is close to 2007 such that the equation is satisfied.

D is the answer.

APRToMBA · Answer

This is a little bit tricky. The best I could immediately come up was to bound it before brute forcing it.

Obviously n < 2007

It's a little less obvious to find the lower bound, but essentially I'm asking the question "how do you maximize S(n) for n < 2007". The answer to that is n=1999. For n = 1999, S(n) = 28 and S(S(n)) = 10. Therefore, our lower bound of n should be 2007 - 28 - 10 = 1969.

I then brute forced every number between 1969 and 2007 and found the following.

n = 1977, 1980, 1983, 2001

nick1816 · Answer

There are 3 possibilities-

1.
  n=0 mod 9  or  9k

As n<2000
S(n) can be 9, 18 or 27

If S(n)= 9 , n= 1989(not possible)
 If S(n)= 18, n= 1980 
If S(n)= 27 , n= 1971(not possible)

2. 
 n=3 mod 9  or  9k+3

As n <2007
S(n)= 3, 12 or 21

If S(n)= 3 , n= 2001 
If S(n)= 12, n= 1992 (not possible)
 If S(n)= 21 , n= 1983

3.
 n= 6 mod 9  or  9k+6

As n<2007
S(n)= 6, 15 or 24

If S(n)= 6 , n= 1995(not possible)
If S(n)= 15, n= 1986 (not possible)
 If S(n)= 24 , n= 1977

Total 4 cases are possible

nick1816 · Answer

You gotta check only 9 cases. n, S(n) or S(S(n) leaves same remainder when divided by 9. As their sum is a multiple of 9, only 3 remainders are possible, that is 3, 6 or 9.

riyu44 · Answer

sum of "n" number is given as n(n+1) / 2 . If we replace S(n) with n(n+1) /2 the equation will be of power n^4 . Hence there can be 4 value of n possible.

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For each positive integer n, let S(n) denote the sum of the digits of

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For each positive integer n, let S(n) denote the sum of the digits of

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