If the positive integer N is a perfect square, which of the following

Thanks for the tips Bunuel. I had never thought about 2 and 3, but they make sense.

On a similar note..
I. The number of factors can be found using the formula = (p+1)(q+1)(s+1).... where p,q,r are the indices

eg1: 25= 5^2 so the number of factors are (2+1) = 3
eg2: 24 = (2^3) * (3^1) so the number of factors are (3+1)(1+1) = 8

Nice tips Thanks Bunuel

Thank you! Bunuel & g4gmat for sharing your tips.

Hi Bunuel!, is there a way of picking numbers to solve this question?
I think that it would be better than trying to remember the rules about perfect squares during the test
Thanks!

Those are useful properties which are worth to remember, even better if you understand why they are right.

Questions about these properties with explanation why they are right:
help-factors-problem-99145.html?hilit=perfect%20square
perfect-square-94700.html?hilit=perfect%20square

As for picking numbers: you can easily prove that III is not always true as soon as you pick appropriate perfect square, say n=2^2=4 --> 4 has 1 (so odd) prime factor, which is 2. For I and II if you try 2-3 perfect squares you'll see that all of them will have the odd number of distinct factors and the odd sum of the distinct factors and though 2-3 examples do not prove that these statement are ALWAYS true you can make educated guess.

The question asks which of the following MUST be true, or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

Hope it helps.

Thank you Bunuel !

I was not aware of 3 and 4.

With reference to point#2, though "The sum of distinct factors of a perfect square is ALWAYS ODD", the vice versa may not be true. Consider the number 2 (factors 1 & 2) and number 8 (factors 1, 2, 4, & 8) -- these are not perfect squares but sum of their distinct factors are odd.

That's correct:

Tips about perfect squares:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

Hope it helps.

Is 0 considered a perfect square?

A perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square.

Since 0=0^2 then 0 is a perfect square. But the properties discussed do not apply to 0.

Hi Bunuel,

In statement 2 you say that the sum of distinct factors of a perfect square is ALWAYS odd but if we consider the perfect square 49 its factors are =7*7*1 and here the distinct factoros of 49 are 7 and 1 which sum to 8 an EVEN number. Can you please help me understand how statement 2 is always true?

Thanks,
Aamir.

Factors of 49 are 1, 7, and 49: 1 + 7 + 49 = 57 = odd.

Let me try to explain it algebraically.
The perfect number A is factored as : a^2 x b^2 x c^2 (the power is always EVEN, for the sake of simplicity, I use the power of 2).
1. The # of distinct factors = 3 x 3 x 3 = 27 : ODD
2. The sum of distinct factors = (a^3 - 1) (b^3 - 1) (c^3 - 1) / [ (a-1)(b-1)(c-1) = (a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)
because a,b,c are different primes then there is at most one even factor among a, b, and c. Let's say a = 2 -> (a^2 + a + 1) = EVEN + EVEN + ODD = ODD
b, c must be ODD -> (b^2 + b + 1) = ODD + ODD + ODD = ODD and (c^2 + c + 1) = ODD + ODD + ODD = ODD.
SO (a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1) = ODD x ODD x ODD = ODD.
3. A = a^2 x b^2 x c^2 if a is EVEN and b, c are ODD.
The # of possible factors of A = 3 x 3 x 3 = 27 = ODD.
The # of possible factors not containing a (so will be ODD) = 3 x 3 = 9 = ODD.
-> The # of possible factors containing a (so will be EVEN) is : 27 - 9 = 18 = ODD - ODD = EVEN.

Hope it helps.

what about for (iii) if N = 1? then 1^2 = 1
therefore 0 even

N = 2 then 2^2 = 4
1, 2, 4
2 even factors

N = 3 then 3^2 = 9
1, 3, 9
0 even factors?

0 is an even integer.

Yes thats what I mean...then isnt 3 true? why is only I and II true?