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

Question

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)}}

KarishmaB · Accepted Answer

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.

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.

Answer (A)

anordinaryguy · Answer

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

So answer is A

Berbatov · Answer

Can you explain that a bit more?

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

eragotte · Answer

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.

nish21in · Answer

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.

nish21in · Answer

in the above post, p1, p2, p3 are prime factors.

BillyZ · Answer

Official solution from Veritas Prep.

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.

jaykhan · Answer

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!

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!

bumpbot · Answer

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

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