GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Nov 2018, 17:48

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • FREE Quant Workshop by e-GMAT!

     November 18, 2018

     November 18, 2018

     07:00 AM PST

     09:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score. November 18th, 7 AM PST
  • How to QUICKLY Solve GMAT Questions - GMAT Club Chat

     November 20, 2018

     November 20, 2018

     09:00 AM PST

     10:00 AM PST

    The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.

2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
User avatar
Joined: 13 Jul 2010
Posts: 139
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post Updated on: 05 Dec 2013, 01:57
3
24
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

64% (01:58) correct 36% (02:24) wrong based on 462 sessions

HideShow timer Statistics

2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

A. 2^22(2^(23)-1)

B. 2^22(2^(24)-1)

C. 2^22(2^(25)-1)

D. 2^23(2^(21)-1)

E. 2^23(2^(22)-1)

Originally posted by gettinit on 21 Nov 2010, 20:34.
Last edited by Bunuel on 05 Dec 2013, 01:57, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Most Helpful Community Reply
Veritas Prep and Orion Instructor
User avatar
B
Joined: 26 Jul 2010
Posts: 309
Re: number properties question  [#permalink]

Show Tags

New post 22 Nov 2010, 12:18
19
19
Hey guys,

GREAT question - thanks for sharing this! This one is a terrific platform for what I call the Three Guiding Principles for Exponents:

1) Find common bases (which usually means you have to factor out bases into prime factors)

2) Multiply (which means that you have to factor out addition/subtraction terms to turn them into multiplication)

3) Find patterns (usually using small numbers to predict how patterns will look with the larger numbers you're given)


Here, I'll use Guiding Principles 2 and 3 (because all the bases are already 2):

All those terms - 2^22 through 2^44 - are going to be too much to carry around, so I'll use a smaller set to try to establish a pattern. Let's say it was just:

2^22 + 2^23 + 2^24 + 2^25

Then we'd need to use Guiding Principle #2 to turn this addition into multiplication, by factoring out the common 2^22 from all terms:

2^22 (1 + 2^1 + 2^2 + 2^3)

2^22 (1 + 2 + 4 + 8)

2^22 (15)

Looking at the answer choices we know we want to have 2^exponent - 1 in there, and we can make the 15 here 16-1, or 2^4 - 1, so we have:

2^22 (2^4 - 1) ----> Started with 4 terms (2^22 through 2^25) and ended up with the 2^22 multiplied by 2^4 - 1


Let's try again, but with five terms to start to see how the pattern goes:

2^22 + 2^23 + 2^24 + 2^25 + 2^26

Factor to:

2^22 (1 + 2 + 2^2 + 2^3 + 2^4)

2^22 (1 + 2 + 4 + 8 + 16)

2^22 (31)

And if we put it in that form as the answer choices we get:

2^22 (2^5 - 1) ----> starting with 5 terms, we end up with 2^22 multiplied by 2^5 - 1


Looking at the pattern, it should become fairly clear that we'll always end up with 2^22 (what we factor) multiplied by 2^an exponent equal to the number of terms we started with, minus 1. So the number of terms that we start with will equal the exponent in 2^x - 1.

Since we'll have 2^22 through 2^44, we have 44-22 + 1 terms (it's an inclusive set), so 23 terms total, and we'll end up with:

2^22 * (2^23 - 1)

And that's answer choice A.
_________________

Brian

Save $100 on live Veritas Prep GMAT Courses and Admissions Consulting

Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Veritas Prep Reviews

General Discussion
SVP
SVP
User avatar
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1855
Concentration: General Management, Nonprofit
GMAT ToolKit User
Re: number properties question  [#permalink]

Show Tags

New post 21 Nov 2010, 21:18
2
gettinit wrote:
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

Please explain methodology.


\(2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)\)

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
\(a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)\)

Thus I think the final answer is B.
Manager
Manager
avatar
S
Joined: 20 Apr 2010
Posts: 217
Location: Hyderabad
WE 1: 4.6 years Exp IT prof
Reviews Badge CAT Tests
Re: number properties question  [#permalink]

Show Tags

New post 22 Nov 2010, 12:06
Answer should be A not B
_________________

I will give a Fight till the End

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
- Bernard Edmonds

A person who is afraid of Failure can never succeed -- Amneet Padda

Don't Forget to give the KUDOS

SVP
SVP
User avatar
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1855
Concentration: General Management, Nonprofit
GMAT ToolKit User
Re: number properties question  [#permalink]

Show Tags

New post 22 Nov 2010, 12:32
Lol, sorry, that was my mistake. :) Answer is A. Thanks Brian!
Manager
Manager
User avatar
Joined: 13 Jul 2010
Posts: 139
Re: number properties question  [#permalink]

Show Tags

New post 22 Nov 2010, 20:28
VeritasPrepBrian wrote:
Hey guys,

GREAT question - thanks for sharing this! This one is a terrific platform for what I call the Three Guiding Principles for Exponents:

1) Find common bases (which usually means you have to factor out bases into prime factors)

2) Multiply (which means that you have to factor out addition/subtraction terms to turn them into multiplication)

3) Find patterns (usually using small numbers to predict how patterns will look with the larger numbers you're given)


Here, I'll use Guiding Principles 2 and 3 (because all the bases are already 2):

All those terms - 2^22 through 2^44 - are going to be too much to carry around, so I'll use a smaller set to try to establish a pattern. Let's say it was just:

2^22 + 2^23 + 2^24 + 2^25

Then we'd need to use Guiding Principle #2 to turn this addition into multiplication, by factoring out the common 2^22 from all terms:

2^22 (1 + 2^1 + 2^2 + 2^3)

2^22 (1 + 2 + 4 + 8)

2^22 (15)

Looking at the answer choices we know we want to have 2^exponent - 1 in there, and we can make the 15 here 16-1, or 2^4 - 1, so we have:

2^22 (2^4 - 1) ----> Started with 4 terms (2^22 through 2^25) and ended up with the 2^22 multiplied by 2^4 - 1


Let's try again, but with five terms to start to see how the pattern goes:

2^22 + 2^23 + 2^24 + 2^25 + 2^26

Factor to:

2^22 (1 + 2 + 2^2 + 2^3 + 2^4)

2^22 (1 + 2 + 4 + 8 + 16)

2^22 (31)

And if we put it in that form as the answer choices we get:

2^22 (2^5 - 1) ----> starting with 5 terms, we end up with 2^22 multiplied by 2^5 - 1


Looking at the pattern, it should become fairly clear that we'll always end up with 2^22 (what we factor) multiplied by 2^an exponent equal to the number of terms we started with, minus 1. So the number of terms that we start with will equal the exponent in 2^x - 1.

Since we'll have 2^22 through 2^44, we have 44-22 + 1 terms (it's an inclusive set), so 23 terms total, and we'll end up with:

2^22 * (2^23 - 1)

And that's answer choice A.


great explanation Brian, and for providing overall methodology on attacking these problems - immensely helpful.
Senior Manager
Senior Manager
User avatar
Joined: 08 Nov 2010
Posts: 341
WE 1: Business Development
GMAT ToolKit User
Re: number properties question  [#permalink]

Show Tags

New post 18 Feb 2011, 23:34
thanks, great explanation. +1
_________________

GMAT Club Premium Membership - big benefits and savings

SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1745
Concentration: Finance
GMAT ToolKit User
Re: number properties question  [#permalink]

Show Tags

New post 04 Dec 2013, 17:53
whiplash2411 wrote:
gettinit wrote:
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

Please explain methodology.


\(2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)\)

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
\(a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)\)

Thus I think the final answer is B.



Dude is that geometric progression formula correct?

Cheers
J :)
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India
Re: number properties question  [#permalink]

Show Tags

New post 04 Dec 2013, 21:04
jlgdr wrote:
whiplash2411 wrote:
gettinit wrote:
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

Please explain methodology.


\(2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)\)

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
\(a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)\)

Thus I think the final answer is B.



Dude is that geometric progression formula correct?

Cheers
J :)


Sum of GP where first term is A, common ratio is R and total number of terms is n is given by
\(Sum = \frac{A(1 - R^n)}{(1 - R)}\)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Intern
avatar
B
Joined: 28 Jan 2017
Posts: 4
Location: India
Concentration: Finance, General Management
WE: Securities Sales and Trading (Mutual Funds and Brokerage)
GMAT ToolKit User
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 22 Mar 2017, 10:05
1
gettinit wrote:
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

A. 2^22(2^(23)-1)

B. 2^22(2^(24)-1)

C. 2^22(2^(25)-1)

D. 2^23(2^(21)-1)

E. 2^23(2^(22)-1)



How about this approach -

A = 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) ... 1
2A = 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) + 2^45 ... 2

2 - 1 gives
A = 2^45 - 2^22 = 2^22 ( 2^23 - 1)
Manager
Manager
avatar
S
Status: GMAT...one last time for good!!
Joined: 10 Jul 2012
Posts: 60
Location: India
Concentration: General Management
GMAT 1: 660 Q47 V34
GPA: 3.5
GMAT ToolKit User Reviews Badge
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 22 Mar 2017, 10:21
dhavalpgk wrote:
gettinit wrote:
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

A. 2^22(2^(23)-1)

B. 2^22(2^(24)-1)

C. 2^22(2^(25)-1)

D. 2^23(2^(21)-1)

E. 2^23(2^(22)-1)



How about this approach -

A = 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) ... 1
2A = 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) + 2^45 ... 2

2 - 1 gives
A = 2^45 - 2^22 = 2^22 ( 2^23 - 1)


Good job! Nice way of going about the problem!
_________________

Kudos for a correct solution :)

GMAT Club Bot
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) = &nbs [#permalink] 22 Mar 2017, 10:21
Display posts from previous: Sort by

2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.