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Does the integer p have an odd number of distinct factors?
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27 Oct 2014, 07:55
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61% (01:19) correct 39% (01:56) wrong based on 147 sessions
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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?
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27 Oct 2014, 09:30
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?
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27 Oct 2014, 09:41
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?
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27 Oct 2014, 09:48
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?
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27 Oct 2014, 11:14
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?
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27 Oct 2014, 21:18
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?
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17 Dec 2015, 07:38
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?
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02 Jul 2019, 09:08
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Re: Does the integer p have an odd number of distinct factors?
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