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Does the integer p have an odd number of distinct factors?

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Does the integer p have an odd number of distinct factors?  [#permalink]

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New post 27 Oct 2014, 06:55
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A
B
C
D
E

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  45% (medium)

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Tough and Tricky questions: Factors.



Does the integer p have an odd number of distinct factors?

(1) p = q^2, where q is a nonzero integer.
(2) p = 2n + 1, where n is a nonzero integer.

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Re: Does the integer p have an odd number of distinct factors?  [#permalink]

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New post 27 Oct 2014, 08:30
Bunuel wrote:

Tough and Tricky questions: Factors.



Does the integer p have an odd number of distinct factors?

(1) p = q2, where q is a nonzero integer.
(2) p = 2n + 1, where n is a nonzero integer.


From 1) if q = 2, p =4 , # distinct factors = 3 -> odd
if q=5 , p = 10, # distinct factors = 4 -> even
not sufficient

From 2) n=> 4, p=>9, #distinct factors = 3 -> odd
n=> 2, p=> 5, factors => 2 even
not sufficient

Together the statements don't make sense to me since p is always even from #1 and always from odd from #2 above. They are contradictory statements.

Any help?
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Re: Does the integer p have an odd number of distinct factors?  [#permalink]

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New post 27 Oct 2014, 08:41
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The strategy I used here is plugging in some numbers.

Lets look at statement 1 first.
q = 1 -> p = 1 Distinct factors: 1
q = 2 -> p = 4 Distinct factors: 1,2,4
q = 3 -> p = 9 Distinct factors: 1,3,9
q = 4 -> p = 16 Distinct factors: 1,4,16
q = 5 -> p = 25 Distinct factors: 1,5,25
q = 6 -> p = 36 Distinct factors: 1,2,3,4,6,9,12,18,36

From the pattern above you can see that p does have an odd number of distinct factors for the numbers plugged in above. Statement 1 is SUFFICIENT. With this you can already eliminate choices B,C and E.

Now, lets look at statement 2. Again plugging in the same numbers for n
n = 1 -> p = 3 Distinct factors: 1,3
n = 2 -> p = 5 Distinct factors: 1,5
n = 3 -> p = 7 Distinct factors: 1,7
n = 4 -> p = 9 Distinct factors: 1,3,9
n = 5 -> p = 11 Distinct factors: 1,11
n = 6 -> p = 13 Distinct factors: 1,13

From the pattern above you can see that p has even and odd number of distinct factors for numbers plugged in above. Statement 2 is NOT SUFFICIENT. Thus, the answer choice is A.
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Re: Does the integer p have an odd number of distinct factors?  [#permalink]

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New post 27 Oct 2014, 08:48
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As per number properties, any square of an integer is integer and has odd number of factors. Thus (1) is sufficient.
From (2), consider 3 and 9. Thus insufficient.
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Does the integer p have an odd number of distinct factors?  [#permalink]

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New post 27 Oct 2014, 10:14
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Initial analysis of the question: Only perfect squares have odd number of distinct factors. Hence, the question is essentially asking whether p is a perfect square

a) P = q^2, and as q is a non zero integer, we can interpret that P will always be a perfect square and that it will have odd number of distinct factors. SUFFICIENT

b) p = 2n + 1, we can substitute n=1, in which case P is not a P.F. but if n=0, or 40 then P = 1, or 81, and is a P.F. Thus, the condition is clearly INSUFFICIENT to make any conclusion based on it.

Thus, answer is A.
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Re: Does the integer p have an odd number of distinct factors?  [#permalink]

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New post 27 Oct 2014, 20:18
As per GMAT CLUB math book :

1) p=q^2

no of different factors of p= 2+1=3 which is odd. hence SUFFICIENT.

2) p=2n+1

taking n=10 ; p=21=3*7
so different factors of p =(1+1)*(1+1)=4 which is even.
taking n=1; p=3 which is odd.
hence its INSUFFICIENT.

Answer must be non other than A
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Re: Does the integer p have an odd number of distinct factors?  [#permalink]

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New post 17 Dec 2015, 06:38
the question basically asks: is p an perfect square?
only perfect squares have an odd number of factors.

statement 1 directly says that p is a perfect square, thus this statement is sufficient.

statement 2 says that p=2n+1. if n=1, then p is 3, a prime number, and thus has only 2 factors.
if n=4, then p=9, which is a perfect square. thus, statement 2 alone is insufficient.
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Re: Does the integer p have an odd number of distinct factors?  [#permalink]

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Re: Does the integer p have an odd number of distinct factors? &nbs [#permalink] 25 Feb 2018, 03:20
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