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Challenge 12 - Q21

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Senior Manager
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Challenge 12 - Q21 [#permalink]

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New post 17 Jun 2006, 02:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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Please explain your choice. I have the OA and the OE, but my answer is different.
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Re: Challenge 12 - Q21 [#permalink]

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New post 17 Jun 2006, 07:23
deowl wrote:
Image

Please explain your choice. I have the OA and the OE, but my answer is different.


from 1, if x = 2, its number of dividers (2= 1 and 2) is smaller than 2 sqrt(x)-1, which is 1.41. if x = 32, yes.

from 2, if x is a prime, again its not.

togather is also not. so E.
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 [#permalink]

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New post 17 Jun 2006, 08:22
Yep - same method. Should be E.

#1. if x=1, f(x)=1 = 1 (nbr of divisors)
if x=2, f(x)=1.82 < 2 (nbr of divisors)
if x=15, f(x)=6.7 > 4 (nbr of divisors)

Not sufficient

#2. if x=97, f(x)=18.7 > 2 (nbr of divisors); and taking
if x=2, f(x)=1.82 < 2 (nbr of divisors)

Not sufficient

f(x)=2*sqrt(x) - 1

:?: Even if we combine, the effect is the same as #2. None of the prime numbers are squares of any integer anyway. Not sufficient.
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Re: Challenge 12 - Q21 [#permalink]

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New post 17 Jun 2006, 08:56
MA wrote:
deowl wrote:
Image

Please explain your choice. I have the OA and the OE, but my answer is different.


from 1, if x = 2, its number of dividers (2= 1 and 2) is smaller than 2 sqrt(x)-1, which is 1.41. if x = 32, yes.
from 2, if x is a prime, again its not.
togather is also not. so E.


little corrections:

from 1, if x = 2, its number of dividers (2 = 1 and 2) is greater than 2 sqrt(2)-1, which is 1.82. if x = 32, the no of divisors is smaller than 2 sqrt(32)-1. so not suff.

from 2, if x is a prime, again its not. x could be 2 or 3 or 5 or 31 or so on.

togather is also not. so E.
Senior Manager
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Joined: 09 Mar 2006
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 [#permalink]

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New post 17 Jun 2006, 09:11
Thanks guys, that's what I got.
Seems just another faulty question.
  [#permalink] 17 Jun 2006, 09:11
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Challenge 12 - Q21

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