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Re: If n and y are positive integers and n represents the number of differ [#permalink]
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Here we go..

n represents the number of different positive factors of y (given)

For Y to be perfect square, n has to be odd.

St1: √n is an odd integer

√n = odd
n = odd^2 = Odd

Clearly sufficient


St2: y = \sqrt{5^{2(n-1)}}

y = 5^(n-1)

n = 2, y = 5 (not a perfect square)
n = 3, y = 25 (perfect square)


Not sufficient.


option A is correct.
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
DesiGmat wrote:
Here we go..

n represents the number of different positive factors of y (given)

For Y to be perfect square, n has to be odd.

St1: √n is an odd integer

√n = odd
n = odd^2 = Odd

Clearly sufficient


St2: y = \sqrt{5^{2(n-1)}}

y = 5^(n-1)

n = 2, y = 5 (not a perfect square)
n = 3, y = 25 (perfect square)


Not sufficient.


option A is correct.


Can you explain why when n is odd then y is a perfect square?
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
Property of perfect square.

Total numbers of prime factors of a perfect square are always odd.

Therefore : A
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
Hi Kuttingchai,

I dont understand your logic here. Please could you explain....

36, which is a perfect square= the factors 1, 2, 3, 4, 6, 9, 12, 18, and 36. The number are prime factors are 2, which is an even number. Please could you explain your logic.



kuttingchai wrote:
Property of perfect square.

Total numbers of prime factors of a perfect square are always odd.

Therefore : A
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
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Hi shriramvelamuri,

I'm pretty sure that what kuttingchai *meant* to say was:

The total number of FACTORS in a perfect square is always odd.

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Re: If n and y are positive integers and n represents the number of differ [#permalink]
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Bunuel wrote:
If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?

(1) √n is an odd integer
(2) \(y = \sqrt{5^{2(n-1)}}\)

Kudos for a correct solution.



I will prefer option A:

Approach is as follows:

from 1: √n is odd integer, which implies n is odd, so y had odd number of factors. This is possible only when y is a perfect square. so sufficient
from 2: y = \sqrt{25^{(n-1)}}[/m], when n = 2, y =5, but when n =5, y = 625 which is perfect square, two different answers NSF

so A
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
1
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Bunuel wrote:
If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?

(1) √n is an odd integer
(2) \(y = \sqrt{5^{2(n-1)}}\)

Kudos for a correct solution.


all perfect squares have an odd number of total factors

Statement 1: sufficient. √n, is odd, so there is an odd number of total factors.
For example √9 = 3. A perfect square such as y=4 has 1,4, and 2 as different positive factors.
√25 = 5. A perfect square such as y=16 has 1,16, 2, 8, and 4 as different positive factors.

Statement 2: \(y = \sqrt{5^{2(n-1)}}\)
\(y = 5^{(n-1)}\)
if n = 2, y = √5. not a perfect square.
if n = 3, y = 25. perfect square.
Insufficient.

Answer: A
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
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Bunuel wrote:
If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?

(1) √n is an odd integer
(2) \(y = \sqrt{5^{2(n-1)}}\)

Kudos for a correct solution.


If a number N is represented as p1^n1*p2^n2....
then number of positive factors of a number = (n1+1)(n2+1)...
So if y is a perfect square, then n1, n2, n3 are all even (since for perfect square, each prime number should be multiplied at least twice). So the number of prime factors = (n1+1)*(n2+2)... will be odd. So the question really is, "is n odd?"
A. If sqrt(n) is odd, since n is positive integer, n is odd. SUFFICIENT.
B. Solving (since y is positive) y = 5^(n-1). So y is 1,5,25,125.... NOT SUFFICIENT.

Answer A.
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
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Bunuel wrote:
If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?

(1) √n is an odd integer
(2) \(y = \sqrt{5^{2(n-1)}}\)

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Correct Answer: A

This difficult question depends on an understanding of the properties of factors of a perfect square. The only way that a number can have an odd number of different total factors is if that number is a perfect square. Consider three numbers that are perfect squares, all of which have an odd number of total factors: 1, 9, and 16. The number 1 has only one unique factor (1); the number 9 has exactly three different factors (1, 3, 9); and 16 has five total factors (1, 16, 2, 8, 4). Statement (1) proves that n must be an odd integer so the information is sufficient to prove that y has an odd number of total unique factors and thus must be a perfect square. Statement (2) is more difficult to assess and requires that several different possible values of n be considered. For instance if n = 1 then y would be equal to 1 and it would indeed be a perfect square. However, if n = 2 then y would be equal to 5 which is not a perfect square. From those two values of n, it is clear that statement (2) is not sufficient and the correct answer is A, statement 1 alone is sufficient.
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
Bunuel wrote:
If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?

(1) √n is an odd integer
(2) \(y = \sqrt{5^{2(n-1)}}\)


n,y: positive integers
n = f(y)

if y is a perf square, then n=odd
ie. y=25, f(25)=5ˆ2=(2+1)=3
ie. y=16, f(16)=2ˆ4=(4+1)=5

(1) √n is an odd integer sufic

if √n=odd, (n positive integer), then n=oddˆ2 which is an odd number

(2) \(y = \sqrt{5^{2(n-1)}}\) insufic

\(y = \sqrt{5^{2(n-1)}}=5ˆ(n-1)\)

n=1: 5ˆ0=1 perfsquare=yes
n=2: 5ˆ(2-1)=5 perfsquare=no

Ans (A)
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
Am I the only one who think 2) automatically answers the question thus should be sufficient?
As we already know Y is a positive integer, then 2) tells us Y is the square root of a certain number, then Y is surely a perfect square!
Pls tell me if I'm wrong.
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
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st8880 wrote:
Am I the only one who think 2) automatically answers the question thus should be sufficient?
As we already know Y is a positive integer, then 2) tells us Y is the square root of a certain number, then Y is surely a perfect square!
Pls tell me if I'm wrong.


If n = 2, then y = 5, and 5 is not a perfect square.
If n = 3, then y = 25, and 25 IS a perfect square.
If n = 4, then y = 125, and 125 is not a perfect square.
etc

Cheers,
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Re: If n and y are positive integers and n represents the number of differ [#permalink]
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