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

It is currently 11 Dec 2019, 11:55

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

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

Find Similar Topics 
Manager
Manager
User avatar
Joined: 13 Jul 2010
Posts: 100
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post Updated on: 08 Mar 2019, 08:07
4
33
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

64% (01:56) correct 36% (02:23) wrong based on 374 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, 21:34.
Last edited by Bunuel on 08 Mar 2019, 08:07, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59674
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 06 Nov 2017, 02:02
3
3
arabella 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 \(2^{23}(2^{(22)}-1)\)


You can solve this question by applying the GP formula.

\(2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} =2^{22}(1+2+2^2+...+2^{22})=\)

Now, again you can apply the formula for GP or just observe the pattern:

\(1 + 2 = 3 =4-1 =2^2 - 1\);
\(1 + 2 + 2^2 = 7 = 8-1=2^3 - 1\);
\(1 + 2 + 2^2 + 2^3 = 15 = 16-1=2^4 - 1\);
...
\(1+2+2^2+...+2^{22} = 2^{23} - 1\).

Thus, \(2^{22}(1+2+2^2+...+2^{22})=2^{22}(2^{23} - 1)\).

Answer: A.

Similar question: https://gmatclub.com/forum/36-126078.html
_________________
Most Helpful Community Reply
Veritas Prep Representative
User avatar
S
Joined: 26 Jul 2010
Posts: 417
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 22 Nov 2010, 13:18
23
20
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

Curriculum Developer, Instructor, and Host of Veritas Prep On Demand

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.

Veritas Prep Reviews
General Discussion
SVP
SVP
User avatar
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1819
Concentration: General Management, Nonprofit
GMAT ToolKit User
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 21 Nov 2010, 22:18
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.
SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1545
Concentration: Finance
GMAT ToolKit User
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 04 Dec 2013, 18:53
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
V
Joined: 16 Oct 2010
Posts: 9869
Location: Pune, India
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 04 Dec 2013, 22:04
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 >
Intern
Intern
avatar
B
Joined: 28 Jan 2017
Posts: 3
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, 11:05
3
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)
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 8302
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 06 Nov 2017, 02:52
arabella 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 \(2^{23}(2^{(22)}-1)\)



a different way and a point to remember.....
\(2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^{n+1}-2\)

so...
\(2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} = (2^1+2^2+......2^{44})-(2^1+2^2+....+2^{21}\)
as per above formula
\((2^1+2^2+......2^{44})-(2^1+2^2+....+2^{21}=(2^{(44+1)}-2)-(2^{(21+1)}-2)=2^{(44+1)}-2-2^{(21+1)}+2=2^{(45)}-2^{(22)}=2^{22}(2^{23}-1)\)
A

you can check why \(2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^{n+1}-2\) at
https://gmatclub.com/forum/2-252919.html
_________________
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2809
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 08 Nov 2017, 17:26
arabella 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 \(2^{23}(2^{(22)}-1)\)


Let’s begin by factoring out 2^22, which is common to each of the terms:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22)

To find an expression for 1 + 2 + 2^2 + … + 2^22, we will use the formula

a^n - 1 = (a - 1)(a^(n - 1) + a^(n - 2) + … + a + 1).

If we let a = 2 and n = 23, we obtain

2^23 - 1 = (2 - 1)(2^22 + 2^21 + … + 2 + 1).

Thus:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22) = 2^22(2^23 - 1)

Answer: A
_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13743
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =  [#permalink]

Show Tags

New post 23 Dec 2018, 09:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =   [#permalink] 23 Dec 2018, 09:35
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  





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