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# If n and y are positive integers and n represents the number

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If n and y are positive integers and n represents the number [#permalink]

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22 Aug 2011, 06:34
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If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?

(1) $$\sqrt{n}$$ is odd number

(2) $$y=\sqrt{5^{2(n-1)}}$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Jul 2013, 01:56, edited 3 times in total.
Edited the question.

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Re: is y a perfect square? [#permalink]

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22 Aug 2011, 10:00
WishMasterUA 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) square(n) is odd number

2) y=square(5^(2(n-1))

I think 'square' means 'square-root'. Considering it true, my explanation

\sqrt{n} is an odd number => n would also be an odd number
=> Sufficient to determine, Y would not have 'even number of factors' and that means it can't a perfect square.

y = \sqrt{5^(2(n-1))} => y= 5^(n-1) => if n is odd than (n-1) would be a perfect square but if n is even, y would not be perfect answer.
=> NOT Sufficient

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Re: is y a perfect square? [#permalink]

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22 Aug 2011, 10:53
anordinaryguy wrote:
=> Sufficient to determine, Y would not have 'even number of factors' and that means it can't a perfect square.

Can you explain that a bit more?

Edit: Found an answer myself: mathforum. org/library/drmath/view/72126.html
Such specific information...

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Re: is y a perfect square? [#permalink]

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22 Aug 2011, 11:20
anordinaryguy wrote:
WishMasterUA 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) square(n) is odd number

2) y=square(5^(2(n-1))

I think 'square' means 'square-root'. Considering it true, my explanation

\sqrt{n} is an odd number => n would also be an odd number
=> Sufficient to determine, Y would not have 'even number of factors' and that means it can't a perfect square.

y = \sqrt{5^(2(n-1))} => y= 5^(n-1) => if n is odd than (n-1) would be a perfect square but if n is even, y would not be perfect answer.
=> NOT Sufficient

For (1) my understanding is different. A perfect square must have event exponents on it's prime factorization, but an odd number of factors. So like you said, root n is odd, meaning Y has an odd number of factors. If a number has an odd number of factors it is a perfect square.

e.g.

1X4
2X2

3 factors

1x64
2x32
4x16
8x8

7 factors

The perfect square always has odd factors.

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Re: is y a perfect square? [#permalink]

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22 Aug 2011, 16:27
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if a number equals = (p1^k)* (p2^l)*(P3^m)....
total no of factors= (k+1)*(l+1)*(m+1)....

If Sqrtroot of n is add => n is also odd.
here n= (k+1)*(l+1)*(m+1)....

now for n to be odd each of k, l,m should be even which implies that y is a perfect square.

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Re: is y a perfect square? [#permalink]

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22 Aug 2011, 16:28
in the above post, p1, p2, p3 are prime factors.

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Re: is y a perfect square? [#permalink]

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22 Aug 2011, 22:42
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WishMasterUA 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) $$\sqrt{n}$$ is an odd number.

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

The question is based on the following concept:
If a number has odd number of factors, it must be a perfect square.
If a number has even number of factors, it cannot be a perfect square.
For why and how, check: http://www.veritasprep.com/blog/2010/12 ... t-squares/

n is the number of positive factors of y.
Question: Is y a perfect square?
Re-state the question as: Is n an odd integer?

Statement 1: $$\sqrt{n}$$ is an odd number.
If $$\sqrt{n}$$ is an odd number, n must be an odd number.
(All powers of an odd number are odd. If a is odd, a^2 is odd, a^3 is odd, a^4 is odd, $$\sqrt{n}$$, if integral, is odd etc.)
Since we know that n is odd, this statement is sufficient.

Statement 2: $$y=\sqrt{5^{2(n-1)}}$$
$$y=5^{n-1}$$
Obviously, the number of factors of y is n. Do we know whether n is odd? No, we don't.
Hence this statement alone is not sufficient.

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

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07 Mar 2017, 18:24
WishMasterUA 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) $$\sqrt{n}$$ is odd number

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

Official solution from Veritas Prep.

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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If n and y are positive integers and n represents the number [#permalink]

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07 Oct 2017, 11:51
Statement 1 says square root of n is odd -> n is also odd and we know that total number of factors of any number can be derived using (n+1)(m+1)... where n & m are the power of any prime factors of a number. (Example: 108 = 2^2*3^3, so the total number of factors is (2+1)*(3+1) = 3*4 = 12 ) Hence, for perfect number such as 4, 9, total number of factors will always be an odd, unlike non perfect numbers, since (even + 1 = odd) -> (2+1 = 3, 6+1 =7). Hence, since statement 1 clearly says y has an odd number of factors, y is definitely a perfect square. Sufficient!

With Statement 2, we can easily put values of n as 1, 2 & 3 and then we can easily derive that with n=2, y is 5 which is not a perfect square and with n=3, y=25 -> clearly a perfect square. Hence, Not sufficient!

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If n and y are positive integers and n represents the number   [#permalink] 07 Oct 2017, 11:51
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