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# value of a

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VP
Joined: 18 May 2008
Posts: 1176

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03 Feb 2009, 04:28

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SVP
Joined: 07 Nov 2007
Posts: 1728
Location: New York

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03 Feb 2009, 05:52
ritula wrote:

I think its B.

product of first 8 postive integers.

1*2*3*4*5*6*7*8
1*2*3*2^2*5*(2*3)*7*(2^3) = a^n =a^6

posible only for a=2

B is sufficient.
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VP
Joined: 17 Jun 2008
Posts: 1474

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03 Feb 2009, 08:09
From stmt1:

1*2*3*(2^2)*5*(2*3)*7*(2^3)
=(2^7)*(3^2)*5*7

Hence, a could be 2 or 3. Insufficient.

From stmt2: a^6 = 64 and a > 1 hence a = 2.
Manager
Joined: 04 Jan 2009
Posts: 229

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03 Feb 2009, 10:30
scthakur wrote:
From stmt1:

1*2*3*(2^2)*5*(2*3)*7*(2^3)
=(2^7)*(3^2)*5*7

Hence, a could be 2 or 3. Insufficient.

From stmt2: a^6 = 64 and a > 1 hence a = 2.

Stmt 1:
A^n = 64. But a could be 2 or a could be 4 hence not sufficient.

stmt 2:
m*a^n = 1x2x3x4x5x6x7x8
So we have 3 unknowns and not enough information.
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tusharvk

VP
Joined: 17 Jun 2008
Posts: 1474

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03 Feb 2009, 10:34
tusharvk wrote:
Stmt 1:
A^n = 64. But a could be 2 or a could be 4 hence not sufficient.

stmt 2:
m*a^n = 1x2x3x4x5x6x7x8
So we have 3 unknowns and not enough information.

Thanks for correction. Answer should be C,
SVP
Joined: 07 Nov 2007
Posts: 1728
Location: New York

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03 Feb 2009, 16:51
scthakur wrote:
tusharvk wrote:
Stmt 1:
A^n = 64. But a could be 2 or a could be 4 hence not sufficient.

stmt 2:
m*a^n = 1x2x3x4x5x6x7x8
So we have 3 unknowns and not enough information.

Thanks for correction. Answer should be C,

It should B
1*2*3*2^2*5*(2*3)*7*(2^3) =m* a^n =m*a^6

2^6*3^2*5*7*2 = m*a^6
even though you have two variables and 1 equation.. there is only on possiblity for a^6 (given a>1) --> 2^6
a=2 and m =3^2*5*7*2
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VP
Joined: 17 Jun 2008
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04 Feb 2009, 03:14
Thanks x2suresh! I realized that I need a break

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Re: value of a &nbs [#permalink] 04 Feb 2009, 03:14
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# value of a

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