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Exponents !

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Senior Manager
Senior Manager
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Joined: 19 Nov 2007
Posts: 477
Followers: 3

Kudos [?]: 51 [0], given: 4

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Exponents ! [#permalink] New post 04 Feb 2010, 17:17
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (02:37) correct 0% (00:00) wrong based on 1 sessions
4 + 4 + 4 + 4 + 3 X 4^2 + 3 X 4^3 + 3 X 4^4 + 3 X 4^5 + 3 X 4^6 + 5^7 =

(A) 47
(B) 48
(C) 4^7 + 5^7
(D) 58
(E) 207

What is the quickest way to solve this :?:

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Manager
Manager
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Joined: 26 May 2005
Posts: 210
Followers: 2

Kudos [?]: 69 [0], given: 1

Re: Exponents ! [#permalink] New post 04 Feb 2010, 17:57
vscid wrote:
4 + 4 + 4 + 4 + 3 X 4^2 + 3 X 4^3 + 3 X 4^4 + 3 X 4^5 + 3 X 4^6 + 5^7 =

(A) 47
(B) 48
(C) 4^7 + 5^7
(D) 58
(E) 207

What is the quickest way to solve this :?:

I would do the below if the answer choices are somewhat close to the sum. I think the answer choices given are wrong. Without much calculations we can know that the sum if more than 207. So could select C.

But if the answer choice are like 4^7 etc .. then would follow the below mentioned approach.

first get the first 4 items sum = 4^2
once you get this the patterns will be similar .. everything is 4^n + 3 * 4^n .. which would be 4^(n+1) .. so the sum of all 4^.... wil be 4^7

C
Senior Manager
Senior Manager
User avatar
Joined: 19 Nov 2007
Posts: 477
Followers: 3

Kudos [?]: 51 [0], given: 4

GMAT ToolKit User
Re: Exponents ! [#permalink] New post 04 Feb 2010, 20:40
chix475ntu wrote:
vscid wrote:
4 + 4 + 4 + 4 + 3 X 4^2 + 3 X 4^3 + 3 X 4^4 + 3 X 4^5 + 3 X 4^6 + 5^7 =

(A) 47
(B) 48
(C) 4^7 + 5^7
(D) 58
(E) 207

What is the quickest way to solve this :?:

Quote:
Without much calculations we can know that the sum if more than 207. So could select C.


Blimey! I sure missed it!

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Re: Exponents !   [#permalink] 04 Feb 2010, 20:40
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