BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Sequence Problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Tue May 11, 2010 12:28 am
Thanked: 1 times

Sequence Problem

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]
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pm
Thanked: 215 times
Followed by:7 members

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!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

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).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Thu Aug 05, 2010 2:39 am

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...
Top
Post new topic Post Reply
• Page 1 of 1