enigma123 wrote:

The infinite sequence Sk is defined as Sk = 10 Sk – 1 + k, for all k > 1. The infinite sequence An is defined as An = 10 An – 1 + (A1 – (n - 1)), for all n > 1. q is the sum of Sk and An. If S1 = 1 and A1 = 9, and if An is positive, what is the maximum value of k + n when the sum of the digits of q is equal to 9?

(A) 6

(B) 9

(C) 12

(D) 16

(E) 18

Guys - any idea how to solve this? I am really struggling without the OA. Therefore your help will be very much appreciated.

For such kind of sequence problems when the formula of \(n_{th}\) term is given it's almost always a good idea to write down first few terms.

Given: \(S_k=10*S_{k-1}+k\) and \(A_n=10*A_{n-1}+A_1-(n-1)\);

\(S_1=1\) and \(A_1=9\);

\(S_2=10*1+2=12\) and \(A_2=10*9+9-(2-1)=98\);

\(S_3=10*12+3=123\) and \(A_3=10*98+9-(3-1)=987\);

\(Q_1=S_1+A_1=1+9=10\) - the sum of the digit of Q is 1;

\(Q_2=S_2+A_2=12+98=110\) - the sum of the digit of Q is 2;

\(Q_3=S_1+A_3=123+987=1,110\) - the sum of the digit of Q is 3;

...

We can see the pattern in \(Q_n\): the sum of its digit equals to \(n\) itself. So the first Q for which the sum of its digit is multiple of 9 is for \(Q_9=S_9+A_9\) --> sum of the digits of \(Q_9\) is 9 --> \(k+n=9+9=18\).

Answer: E.

P.S. I wonder whether the question supposed to ask the minimum value of k + n when the sum of the digits of q is equal to 9, though anyway as 18 is the the largest value from among the answer choices it's still the right one.

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