Exponents ! : GMAT Problem Solving (PS)
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# Exponents !

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Senior Manager
Joined: 19 Nov 2007
Posts: 470
Followers: 4

Kudos [?]: 196 [0], given: 4

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04 Feb 2010, 17:17
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Question Stats:

100% (02:37) correct 0% (00:00) wrong based on 1 sessions

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$$4 + 4 + 4 + 4 + 3 X 4^2 + 3 X 4^3 + 3 X 4^4 + 3 X 4^5 + 3 X 4^6 + 5^7$$ =

(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
Joined: 26 May 2005
Posts: 208
Followers: 2

Kudos [?]: 119 [0], given: 1

Re: Exponents ! [#permalink]

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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
Joined: 19 Nov 2007
Posts: 470
Followers: 4

Kudos [?]: 196 [0], given: 4

Re: Exponents ! [#permalink]

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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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# Exponents !

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