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# integral part of x = an integer n such that n <= x <

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Manager
Joined: 25 Jan 2004
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integral part of x = an integer n such that n <= x < [#permalink]

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01 Feb 2004, 03:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

[x] = integral part of x = an integer n such that n <= x < (n+1)

Find [1/3] + [2/3] + [(2^2)/3] + [(2^3)/3] + ... + [(2^1000)/3].

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GMAT Instructor
Joined: 07 Jul 2003
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04 Feb 2004, 02:40
(2^1001-2)/3 - 500

Can you confirm?
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Manager
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04 Feb 2004, 22:24
I dont have the answer, but the following is what I've got

[(2^n)/3] = (-1/2)+(1/6)*(-1)^n +(1/3)*2^n

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Director
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04 Feb 2004, 23:10
AkamaiBrah wrote:
(2^1001-2)/3 - 500

Can you confirm?

I got (2^1001) / 3.

Could you explain why you are subtracting 500.

Kudos [?]: 106 [0], given: 0

Manager
Joined: 25 Jan 2004
Posts: 92

Kudos [?]: 2 [0], given: 0

Location: China

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05 Feb 2004, 01:12
BTW, the answer should be an integer. But 2^1001/3 is not.

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GMAT Instructor
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Kudos [?]: 240 [0], given: 0

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Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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05 Feb 2004, 03:03
Zhung Gazi wrote:
I dont have the answer, but the following is what I've got

[(2^n)/3] = (-1/2)+(1/6)*(-1)^n +(1/3)*2^n

If you look at your answer closely, you will see that it is the same as mine.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Kudos [?]: 240 [0], given: 0

GMAT Instructor
Joined: 07 Jul 2003
Posts: 769

Kudos [?]: 240 [0], given: 0

Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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05 Feb 2004, 04:47
AkamaiBrah wrote:
(2^1001-2)/3 - 500

Can you confirm?

I got (2^1001) / 3.

Could you explain why you are subtracting 500.

Simplest way to solve this is to assume geometry series of 1001 terms.

If you examine terms, you will notice that rounding every odd and even term subtracts 1/3 and 2/3 respectively. Hence, the answer is the sum of the series or 1/3*(2^1001-1) less the sum of the rounding 500(1/3+2/3) + 1/3 which simplifies to the answer I gave.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Kudos [?]: 240 [0], given: 0

Director
Joined: 03 Jul 2003
Posts: 651

Kudos [?]: 106 [0], given: 0

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05 Feb 2004, 09:12
AkamaiBrah wrote:
Can you confirm?
I got (2^1001) / 3.

Could you explain why you are subtracting 500.

Simplest way to solve this is to assume geometry series of 1001 terms.

My Bad, I didn't read the question stem properly
I've interpreted [X] as just ().

[x] = integral part of x = an integer n such that n <= x < (n+1)
is the key !

Thanks Akamai

Kudos [?]: 106 [0], given: 0

05 Feb 2004, 09:12
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# integral part of x = an integer n such that n <= x <

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