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Tough Sequences Question

by mba_aspirant911 » Sat Dec 04, 2010 1:14 am
Note: the k and n are subscripts in the question below

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

I don't know the OA. I took a fairly cumbersome approach of calculating An and Sk and summing them upto to find one that has the sum of its digits equal to 9. Giving n = 6 and k = 6 and therefore n + k = 12. Theoretically, it is possible to go upto any n or k.

I seriously don't know how one can do this question in 2 minutes..a general approach to a problem such as this would be greatly appreciated. Thanks.
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by Bharat » Sat Dec 04, 2010 11:34 am
Hi mba_aspirant911,

the faster approach is solving a few terms of the series:
S1: 1
S2: 12
S3: 123

A1: 9
A2: 98
A3: 987


Q(1,1): 10 = S1+A1
Q(2,2): 110 = S2+A2
Q(3,3): 1110 = S3+A3
Q(1,2): 99 = (S1 + A2)
Q(2,1): 21 = (S2 + A1)

Looking at the value of Q; Sum of Q's digit will be maximum when both n & k are equal

Hence digits of Q will have a total of 9 for Q(9,9) -> S9 + A9 -> max (k+n) = 9+9 = 18

Let me know if you have question or suggestion.
Regards,
Bharat.
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by diebeatsthegmat » Mon Dec 06, 2010 7:05 pm
Bharat wrote:Hi mba_aspirant911,

the faster approach is solving a few terms of the series:
S1: 1
S2: 12
S3: 123

A1: 9
A2: 98
A3: 987


Q(1,1): 10 = S1+A1
Q(2,2): 110 = S2+A2
Q(3,3): 1110 = S3+A3
Q(1,2): 99 = (S1 + A2)
Q(2,1): 21 = (S2 + A1)

Looking at the value of Q; Sum of Q's digit will be maximum when both n & k are equal

Hence digits of Q will have a total of 9 for Q(9,9) -> S9 + A9 -> max (k+n) = 9+9 = 18

Let me know if you have question or suggestion.
Regards,
Bharat.
can you please explain more in details, please? i dun understand much...how come you find A2 and A3 and Q2 and Q3?
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by Rahul@gurome » Mon Dec 06, 2010 8:00 pm
diebeatsthegmat wrote:
Bharat wrote:Hi mba_aspirant911,

the faster approach is solving a few terms of the series:
S1: 1
S2: 12
S3: 123

A1: 9
A2: 98
A3: 987


Q(1,1): 10 = S1+A1
Q(2,2): 110 = S2+A2
Q(3,3): 1110 = S3+A3
Q(1,2): 99 = (S1 + A2)
Q(2,1): 21 = (S2 + A1)

Looking at the value of Q; Sum of Q's digit will be maximum when both n & k are equal

Hence digits of Q will have a total of 9 for Q(9,9) -> S9 + A9 -> max (k+n) = 9+9 = 18

Let me know if you have question or suggestion.
Regards,
Bharat.
can you please explain more in details, please? i dun understand much...how come you find A2 and A3 and Q2 and Q3?
Sk = 10*S(k-1)+k.
S1 = 1
S2 =10*S1 + 2 = 10*1 + 2 = 12.
S3 = 10*S2 + 3 = 10*12+3 = 123.

An = 10*A(n-1) + (A1 - (n-1)).
A1 = 9.
A2 = 10*A1 + (A1 - (n-1)) = 10*9 + (9 - 1) = 98.
A3 = 10*A2 + (9 - 2) = 980+7 = 987.

q = Sk + An.
Note that q(1) = S1+A1 = 1+9 = 10. (sum of digits of q is 1).
q(2) = S2+A2 = 12 + 98 = 110. (sum of digits of q is 2).
q(3) = S3+A3 = 123+987 = 1110. (sum of digits of q is 3)
So for q(n) = Sn + An = 111...(1 is n times)0. (sum of digits of q is n).
So if sum of digits of q is 9, q can be q(9) = S9 + A9.
So a value of k+n for sum of digits of q being 9 can be 9+9 = 18.
Since 18 is the largest value from among the options, go with it.

The correct answer is e).
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by diebeatsthegmat » Tue Dec 07, 2010 9:50 am
Rahul@gurome wrote:
diebeatsthegmat wrote:
Bharat wrote:Hi mba_aspirant911,

the faster approach is solving a few terms of the series:
S1: 1
S2: 12
S3: 123

A1: 9
A2: 98
A3: 987


Q(1,1): 10 = S1+A1
Q(2,2): 110 = S2+A2
Q(3,3): 1110 = S3+A3
Q(1,2): 99 = (S1 + A2)
Q(2,1): 21 = (S2 + A1)

Looking at the value of Q; Sum of Q's digit will be maximum when both n & k are equal

Hence digits of Q will have a total of 9 for Q(9,9) -> S9 + A9 -> max (k+n) = 9+9 = 18

Let me know if you have question or suggestion.
Regards,
Bharat.
can you please explain more in details, please? i dun understand much...how come you find A2 and A3 and Q2 and Q3?
Sk = 10*S(k-1)+k.
S1 = 1
S2 =10*S1 + 2 = 10*1 + 2 = 12.
S3 = 10*S2 + 3 = 10*12+3 = 123.

An = 10*A(n-1) + (A1 - (n-1)).
A1 = 9.
A2 = 10*A1 + (A1 - (n-1)) = 10*9 + (9 - 1) = 98.
A3 = 10*A2 + (9 - 2) = 980+7 = 987.

q = Sk + An.
Note that q(1) = S1+A1 = 1+9 = 10. (sum of digits of q is 1).
q(2) = S2+A2 = 12 + 98 = 110. (sum of digits of q is 2).
q(3) = S3+A3 = 123+987 = 1110. (sum of digits of q is 3)
So for q(n) = Sn + An = 111...(1 is n times)0. (sum of digits of q is n).
So if sum of digits of q is 9, q can be q(9) = S9 + A9.
So a value of k+n for sum of digits of q being 9 can be 9+9 = 18.
Since 18 is the largest value from among the options, go with it.

The correct answer is e).
thanks, i got it.... misread Sk into Sn :(
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by diebeatsthegmat » Tue Dec 07, 2010 9:52 am
Rahul@gurome wrote:
diebeatsthegmat wrote:
Bharat wrote:Hi mba_aspirant911,

the faster approach is solving a few terms of the series:
S1: 1
S2: 12
S3: 123

A1: 9
A2: 98
A3: 987


Q(1,1): 10 = S1+A1
Q(2,2): 110 = S2+A2
Q(3,3): 1110 = S3+A3
Q(1,2): 99 = (S1 + A2)
Q(2,1): 21 = (S2 + A1)

Looking at the value of Q; Sum of Q's digit will be maximum when both n & k are equal

Hence digits of Q will have a total of 9 for Q(9,9) -> S9 + A9 -> max (k+n) = 9+9 = 18

Let me know if you have question or suggestion.
Regards,
Bharat.
can you please explain more in details, please? i dun understand much...how come you find A2 and A3 and Q2 and Q3?
Sk = 10*S(k-1)+k.
S1 = 1
S2 =10*S1 + 2 = 10*1 + 2 = 12.
S3 = 10*S2 + 3 = 10*12+3 = 123.

An = 10*A(n-1) + (A1 - (n-1)).
A1 = 9.
A2 = 10*A1 + (A1 - (n-1)) = 10*9 + (9 - 1) = 98.
A3 = 10*A2 + (9 - 2) = 980+7 = 987.

q = Sk + An.
Note that q(1) = S1+A1 = 1+9 = 10. (sum of digits of q is 1).
q(2) = S2+A2 = 12 + 98 = 110. (sum of digits of q is 2).
q(3) = S3+A3 = 123+987 = 1110. (sum of digits of q is 3)
So for q(n) = Sn + An = 111...(1 is n times)0. (sum of digits of q is n).
So if sum of digits of q is 9, q can be q(9) = S9 + A9.
So a value of k+n for sum of digits of q being 9 can be 9+9 = 18.
Since 18 is the largest value from among the options, go with it.

The correct answer is e).
thanks, i got it.... misread Sk into Sn :(
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