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

Redeem

Exponent pls help

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Community Manager
Posts: 1049
Joined: Sun Apr 06, 2008 5:15 pm
Location: Pittsburgh, PA
Thanked: 113 times
Followed by:27 members
GMAT Score:710

by dmateer25 » Fri Apr 10, 2009 8:30 am
2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

2(2+2+2^2+2^3+2^4+2^5+2^6+2^7)
2^2(2+2+2^2+2^3+2^4+2^5+2^6)
2^3(2+2+2^2+2^3+2^4+2^5)
2^4(2+2+2^2+2^3+2^4)
2^5(2+2+2^2+2^3)
2^6(2+2+2^2)
2^7(2 + 2)
2^8(2)
2^9
Top
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Fri Apr 10, 2009 9:38 am
Thanked: 1 times

by milqman » Fri Apr 10, 2009 10:37 am
That solution is actually not quite right.

There is an error in the first line of work shown (pulling a 2 out would not leave you with 2's on the inside like that).

Later on the 2^8 line: 2^7*(2+2) does not equal 2^8

I think the easiest way to do this is to evaluate the first handful, then factor out of the remainder and evaluate again.

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

2+2+4+8+16 + (2^4*(2+2^2+2^3+2^4))

32 + (16*(30))

32 + 480

512

2^9

Does that make sense?
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Wed Aug 27, 2008 8:35 am

by ramandeepn » Fri Apr 10, 2009 10:44 am
Thank you so much. Now i understand it... :)
Top
Master | Next Rank: 500 Posts
Posts: 353
Joined: Sat Jan 20, 2007 1:29 am
Location: Italy
Thanked: 7 times
GMAT Score:720

by mjjking » Fri Apr 10, 2009 10:44 am
There are many ways to reach the solution, but I agree that grouping them and multiplying is the easiest one.
Beat The GMAT - 1st priority
Enter a top MBA program - 2nd priority
Loving my wife: MOST IMPORTANT OF ALL!

REAL THING 1 (AUG 2007): 680 (Q43, V40)
REAL THING 2 (APR 2009): 720 (Q47, V41)
Top
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 am
Thanked: 82 times
Followed by:9 members
GMAT Score:720

by maihuna » Fri Apr 10, 2009 11:07 am
the first time when i saw this question in my test itself I was kind of taken by surprise...but in a moment i found this pattern:

2+2 = 2^2
2^2 + 2^2 = 2^3
2^3 + 2^3= 2^4

r u able to see a pattern, previous two terms equate to next term in series...

that way the last two will be : 2^8 + 2^8 = 2^9
Top
User avatar
Community Manager
Posts: 1049
Joined: Sun Apr 06, 2008 5:15 pm
Location: Pittsburgh, PA
Thanked: 113 times
Followed by:27 members
GMAT Score:710

by dmateer25 » Fri Apr 10, 2009 1:45 pm
milqman wrote:That solution is actually not quite right.

Later on the 2^8 line: 2^7*(2+2) does not equal 2^8


2^7 *(4) = 2^8 *(2)

I just showed it as 2^7 *(2+2)

Pulling out the 2 leaves you with 2^8 * (1+1) = 2^8 * (2)
Top
Post new topic Post Reply
• Page 1 of 1