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is y a perfect square

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is y a perfect square

by oldnass » Wed Aug 25, 2010 11:02 am
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 integer
2. y=sqrt[5^[2(n-1)]]

Please help!
OA is A

My reasoning:
1 is insufficient:
Let n = 9, so sqrt(n) = sqrt(9) = 3 is an odd integer. So then y has 9 different positive factors, so say y = 1*2*3*4*5*6*7*8*9, which is not a perfect square.
But if n = 1, so sqrt(1) = 1 is also an odd integer. So then y has 1 factor 1, so y = 1 is a perfect square.

2 is insufficient:
No restrictions on n.

Together they are sufficient:
y = 5^(n-1) so when sqrt( n) is odd then, n is also odd, so y must be a perfect square.

Where am I wrong?
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by 4GMAT_Mumbai » Wed Aug 25, 2010 11:54 am
oldnass wrote: My reasoning:
1 is insufficient:
Let n = 9, so sqrt(n) = sqrt(9) = 3 is an odd integer. So then y has 9 different positive factors, so say y = 1*2*3*4*5*6*7*8*9, which is not a perfect square.
But if n = 1, so sqrt(1) = 1 is also an odd integer. So then y has 1 factor 1, so y = 1 is a perfect square.

Where am I wrong?
Statement 1:

sqrt(n) is an odd integer. This implies that n should also have been an odd integer.

Only perfect square numbers have odd number of 'distinct / different' factors.

For example, 36 is a perfect square. It has 9 factors (1,2,3,4,6,9,12,18,36). 24 is a non perfect square. It has 8 factors (1,2,3,4,6,8,12,24)

In general, if y = (2 ^ a) times (3 ^ b) times (5 ^ c) so on and so forth (y has been expressed as a product of some power of its prime factors); then n = (a+1) times (b+1) times (c+1) and so on ...

You will realize that for perfect squares; all of a,b,c ... will be even. Hence, all of (a+1), (b+1), (c+1) will be odd and therefore n will also be odd.

That is the way to think about the 1st statement. Hope this helps.

Thanks.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by Rich@VeritasPrep » Wed Aug 25, 2010 12:59 pm
Hey oldnass,

The error in your explanation of Statement 1 comes from the fact that y = 1*2*3*4*5*6*7*8*9 does not have 9 factors. It has many more than that. Remember, 1*2*3 counts as a factor, as do 4*5*6, 4*8*6, or any other like combination. It's not just the individual numbers 1 through 9.

As 4GMAT_Mumbai pointed out, the correct way to view this problem is in terms of the perfect-square factor property. This simply states that the only numbers that have an odd number of factors are perfect squares. (By the way, this is because when you try to split up the factor pairs, the square root will be by itself. For example, 16 has an odd number of factors, because 1 pairs with 16 and 2 pairs with 8, but the 4 will be by itself, thus resulting in an odd number of factors.)

As such, if you look closely at the prompt, it can really be boiled down to the following question: "Is n odd?"

This is because they're asking if y is a perfect square. They tell you that n represents the number of factors y has. Well, if y is a perfect square, then n will be odd.

Statement (1): If sqrt(n) is odd, then n is odd. SUFFICIENT

Statement (2): n may or may not be odd. INSUFFICIENT

Make sense?
Rich Zwelling
GMAT Instructor, Veritas Prep
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