Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.

The answer is D: EACH statement ALONE is sufficient.

This is from one of the Manhattan CAT exams. I understand why the 1st statement is sufficient. However, I don't understand the reasoning behind the 2nd statement. It seems like the 2nd statement should be insufficient. Please simplify the explanation.

Hey agalstia, this is two questions in a row where your question is just "I don't get it." You're going to have to be more specific than that.

I'll give this a shot, though. First of all, unless I knew the relevant fact about perfect squares ahead of time, or had stroke of genius, I would never use theory to consider S2. I would just check whether any perfect square had an even number of odd factors. I'd be sure to check numbers likely to give different results, perhaps 1, the squares of primes as well as of composites, and the squares of evens as well as of odds.

I don't mind doing a little theory here though, since the clock isn't running.

Here's a rough proof that the sum of the distinct factors of every perfect square is odd.

I. A perfect square has an even number of each of its prime factors. (See pages 122-124 of your Number Properties guide.)
(From I) II. There are an odd number of possibilities for the number of occurrences of each odd prime factor of a perfect square. (See page 130 of your Number Properties guide.)
(From II) III. A perfect square has an odd number of distinct odd factors.
(From III) IV. The sum of the distinct odd factors of a perfect square is odd.
V. The sum of the distinct even factors of a perfect square is even.
(From IV and V) VI. The sum of the distinct odd factors and the distinct even factors of a perfect square is odd.

That's gonna take more than two minutes. It sure took me longer than two minutes to figure out how to explain it.

For instance, when I take the square root of an odd perfect square like 25, the two factors are 5 and 5. Thus, there is only 1 distinct factor, which is 5, and it is odd.

When I take the square root of an even perfect square, like 16, it is 16=2x2x2x2. There is only 1 distinct factor and it is even.

When I factor out a number such as 35, for example, the factors are 5 and 7. The sum of 5 and 7 is 12, which is even.

Due to this inconsistency, I said that the 2nd statement is Insufficient.

Hi,

In case of 25 (5^2), distinct factors are three and not one These three factors are: 1, 5 and 25.

Sum of these factors comes out as odd.

So This is contrary to statement (2)


In case of 16: ((2^4)^2) = 2^8

here total factors are 9.

Except "1", all are even.

Sum of these factors has to be odd.

And please note that distinct factors are 9 in No. and are as follows:


1, 2, 4, 8, 16, 32, 64, 128, 256.

So, once again we find that

This is contrary to statement (2)

Check 9: 3^2

distinct factors: 1, 3, 9.

sum: odd.

So we can say that (2) is sufficient (to tell us that N is not a perfect square)

Thanks for the explanation. Remember, prime factors are different from factors. In general, think of it this way: prime factors are building blocks of larger numbers. Anything you can build from the prime factors in n's prime box will be a factor of n..

_________________
Tim Sanders
Manhattan GMAT Instructor

I still not get why (1) is sufficient

Pick 6 --> 4 distinct factors 1,2,3,6 but 6 is not a perfect square!!

I choose E but it is incorrect not sure why???

I think I got it now
Please correct me if i am misunderstood
Every Perfect square For example 4, 9 , 25

4 --> 1,2,4
9--> 1,3,9
25--> 1,5,25

all of them has an odd number of distinct factors right
So (1) is sufficient to answer that N is not a perfect square

Also
1+2+4 = 7
1+3+9 = 13
1+5+25 = 31

All of them sums up to odd number
So (2) is sufficient to answer that N is not a perfect square

What do u think?

yep. i think i answered three threads on this same question in the last hour. we're gonna have to merge these. make sure to search! save yourself some time.