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Sequence Problem

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by mba1986 » Tue Aug 10, 2010 11:00 pm
S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk-1 + 2(10k-1). If p is the sum of the first 30 terms of S, what is the eleventh digit of p, counting right to left from the units digit?

A) 1
B) 2
C) 4
D) 6
E) 9

The way shown is far too time consuming. Any help would be greatly appreciated :D.

[spoiler]Source:Mgmat Question Bank[/spoiler]
[spoiler]Answer: C[/spoiler]
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by kvcpk » Wed Aug 11, 2010 12:00 am
arjunsekhar wrote:S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk-1 + 2(10k-1). If p is the sum of the first 30 terms of S, what is the eleventh digit of p, counting right to left from the units digit?

A) 1
B) 2
C) 4
D) 6
E) 9

The way shown is far too time consuming. Any help would be greatly appreciated :D.

[spoiler]Source:Mgmat Question Bank[/spoiler]
[spoiler]Answer: C[/spoiler]
We need the 11th digit of the series:
2+22+222+2222+....

2
22
222
2222
22222
....

2.................2
---------------------------
............41975308642

Let us backtrack the sum
Last digit will be 2
2nd last digit will be 2+2 = 4
3rd Last digit will be 2(3) = 6
4th Last digit will be 2(4) = 8
5th last digit will be 2(5) = 0 -> carry 1
6th will be 2(6)+1 = 12+1 =3 -> carry1
7th will be 2(7) +1 = 14+1 =5 -> carry 1
..
..
10th will be 2(10)+1 = 21 = 1 -> carry 2
11th will be 2(11) +2 = 24 -> 4 is the 11th digit.

Hope this helps!!
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by Stuart@KaplanGMAT » Wed Aug 11, 2010 12:06 am
arjunsekhar wrote:S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk-1 + 2(10k-1). If p is the sum of the first 30 terms of S, what is the eleventh digit of p, counting right to left from the units digit?

A) 1
B) 2
C) 4
D) 6
E) 9

The way shown is far too time consuming. Any help would be greatly appreciated :D.

[spoiler]Source:Mgmat Question Bank[/spoiler]
[spoiler]Answer: C[/spoiler]
Hmm.. if you plug in k=2, you get:

S(2) = S(1) + 2(20 - 1) = 1 + 2(19) = 1 + 38 = 39, which isn't 22.

Similarly,

S(3) = S(2) + 2(30 - 1) = 22 + 2(29) = 22 + 58 = 70, which isn't 222.

If you accept that S(2) = 39, then:

S(3) = S(2) + 2(30 - 1) = 39 + 2(29) = 39 + 58 = 97 (also not 222).

Have you reproduced the entire question?

Note: I'm assuming that you just left out the brackets and your equation is:

S(k) = S(k-1) + 2(10k-1).
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by likithae » Wed Aug 11, 2010 1:17 am
kvcpk wrote:
arjunsekhar wrote:S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk-1 + 2(10k-1). If p is the sum of the first 30 terms of S, what is the eleventh digit of p, counting right to left from the units digit?

A) 1
B) 2
C) 4
D) 6
E) 9

The way shown is far too time consuming. Any help would be greatly appreciated :D.

[spoiler]Source:Mgmat Question Bank[/spoiler]
[spoiler]Answer: C[/spoiler]
We need the 11th digit of the series:
2+22+222+2222+....

2
22
222
2222
22222
....

2.................2
---------------------------
............41975308642

Let us backtrack the sum
Last digit will be 2
2nd last digit will be 2+2 = 4
3rd Last digit will be 2(3) = 6
4th Last digit will be 2(4) = 8
5th last digit will be 2(5) = 0 -> carry 1
6th will be 2(6)+1 = 12+1 =3 -> carry1
7th will be 2(7) +1 = 14+1 =5 -> carry 1
..
..
10th will be 2(10)+1 = 21 = 1 -> carry 2
11th will be 2(11) +2 = 24 -> 4 is the 11th digit.

Hope this helps!!
thank you........can you explain me in another approach please...
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