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Question from Challenge 21

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Question from Challenge 21 [#permalink] New post 19 Mar 2006, 17:16
Please try to explain your answers.

If x is a positive Integer, is the number of its divisors smaller than 2*sqrt(x)-1?

(1) x is not a square of an integer
(2) x is prime
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Re: Question from Challenge 21 [#permalink] New post 19 Mar 2006, 17:38
Praetorian wrote:
Please try to explain your answers.

If x is a positive Integer, is the number of its divisors smaller than 2*sqrt(x)-1?

(1) x is not a square of an integer
(2) x is prime


suppose, n = no of divisors

(I) x could be 1 or 3 or 6 or so on.
=> if x = 1, n = 1 and 2*sqrt(x)-1 = 1.
so n = 2*sqrt(x)-1.
=> if x = 3, n = 2 (1 and 3) and 2*sqrt(3)-1 = 1.8 approx.
so, n>2*sqrt(x)-1.
=> if x = 6, n = 4 (1, 2, 3 and 6) and 2*sqrt(6)-1 = 4.4 approx. divisors<2*sqrt(x)-1.
so not sufficient.

(II) if x is a prime, n is always 2 but 2*sqrt(x)-1 greater or smaller than n. so not sufficient.

from i and ii, togather, also not sufficient. so E.
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 [#permalink] New post 19 Mar 2006, 19:25
1) x is not square of an integer.

x = 3, # of divisors = 2 (1,3)
2*sqrt(x)-1 = 2.46
# of divisors smaller than 2*sqrt(x)-1

x = 11, # of divisors = 2 (1,11)
2*sqrt(x)-1 = 5.6
# of divisors smaller than 2*sqrt(x)-1

x = 8, # of divisors = 4 (1,2,4,8)
2*sqrt(x)-1 = 4.65
# of divisors smaller than 2*sqrt(x)-1

Sufficient.

(2) x is prime
If x is prime, it will always have two divisors, 1 and x.

x = 2, 2*sqrt(x)-1 = 1.8 (less than number of divisors)
x = 5, 2*sqrt(x)-1 = 3.4 (more than number of divisors)

Insufficient.

I go with A
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Re: Question from Challenge 21 [#permalink] New post 19 Mar 2006, 21:03
i guess we cannot take 1 because it is a square but we can use 2. if so,
if x = 2, n = 2 (1 and 2) and 2*sqrt(2)-1 = 1.83 approx.
so, n>2*sqrt(x)-1.

Professor wrote:
suppose, n = no of divisors

(I) x could be 1 or 3 or 6 or so on.

=> if x = 1, n = 1 and 2*sqrt(x)-1 = 1.
so n = 2*sqrt(x)-1.

=> if x = 3, n = 2 (1 and 3) and 2*sqrt(3)-1 = 2.46 approx. so, n>2*sqrt(x)-1.

=> if x = 6, n = 4 (1, 2, 3 and 6) and 2*sqrt(6)-1 = 3.9 approx. divisors<2*sqrt(x)-1.
so not sufficient.

(II) if x is a prime, n is always 2 but 2*sqrt(x)-1 greater or smaller than n. so not sufficient.

from i and ii, togather, also not sufficient. so E.


from statement i, [the following table (calculated in XL format) would be more useful].

if x = 2, n = 2, 2sqrt(x)-1 = 1.83
if x = 3, n = 2, 2sqrt(x)-1 = 2.46
if x = 8, n = 4, 2sqrt(x)-1 = 4.66
if x = 11, n = 2, 2sqrt(x)-1 = 5.63
if x = 12, n = 6, 2sqrt(x)-1 = 5.93

from above when x = 2, and 12, n> 2sqrt(x)-1 and when x = 3, 8, and 11, n<2sqrt(x)-1. this proovs that A is out.

from st ii,
if x = 8, n = 4, 2sqrt(x)-1 = 4.66
if x = 12, n = 6, 2sqrt(x)-1 = 5.93

from above when x = 8, n< 2sqrt(x)-1 and when x = 12, n>2sqrt(x)-1. this proovs that A is out. this also rules out B.

togather also we cannot say which one is greater. so E.
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 [#permalink] New post 20 Mar 2006, 04:43
same method of substituing numbers --> same answer E.
  [#permalink] 20 Mar 2006, 04:43
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Question from Challenge 21

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