If the positive integer N is a perfect square, which of the following must be true?

I. The number of distinct factors of N is odd.
II. The sum of the factors of N is odd.
III. The number of distinct prime factors of N is even.

A) I only
B) II only
C) I and II
D) I and III
E) I, II and III

I have a question about the inclusion of the word "distinct" in statements I and III, but not II. Why is this? One would think that the lack of the word distinct in statement II implies that for a number such as 9, the sum of factors would be (1 + 9 + 3 + 3) where as the sum of distinct factors would be (1 + 9 + 3).

Thanks in advance,
Mike

Not sure about distinct word in II statement
but answer is E.

Can you please post the original answer?

I came across the same problem in one of my MGMAT's.

The correct answer is C; can we have a little discussion on this before I post the given explanation.

How do we approach these kind of problems - where do we start, what numbers should we plug in, how do we eliminate the incorrect options, etc ??

Any easier way to solve these kind of problems? I guessed on this one and moved on.

the positive integer N is a perfect square say: 4 and then 36

N=4

factors: 1,2,4 -1, -2, -4

N=36

1, 2, 2*3, 2*3^2, (2^2*3), (2^2)*(3^2) and -1, -2, -2*3, -2*3^2,
-(2^2*3), -(2^2)*(3^2)

This gives only III as correct which is not in any option.
Please comment on my explanation.

@vishalsahdev03
you are not suppose to consider negative number factors..!

I took 16 and 9 as exmple:

16 - 1,2,4,8,16

==> number of distinct factor = 5

9 - 1,3,9 ==> 3 factors
similary ==>
36 - 1,2,3,4,6,9,12,18,36 --> 9

so Ist statement is true.

II)
sum of factor:-
16 :=> 1,4,4 ==> 9
9 :==> 1,3,3 ==> 7
you know it is perfect square
so N = x * x
sum of two same number is always ==> even
1 is always a factor ==> sum ==> even + 1 = odd

so Its True.

III)
Number of distinct PRIME factor:-
16 :== 2 ==> only one prime factor ==> odd
so III) --> not necessary to be true.

Hence, C is the answer.

Please provide reasoning as in why negative factors should not be considered, the question does not provide any such condition.

hi to every one

Carefully read the question.



-4 is not a perfect square nor is it positive.

Here's the answer. took me 2 mins to solve it.


How I is true:

n= 4
1,2,4
Distinct numbers - 3

n= 16
1,2,4,8,16
Distinct numbers - 5

You will always have one number that is SQUARED. Therefore, making that number DISTINCT and in this case odd.

How II is true
n= 4
1,2,4
1+2+4=7

n= 9
1,3,9
1+3+9=13

n=25
1,5,25
1+5+25=31

You'll notice a pattern where if the perfect square is odd, it's number + 1 is even while the summation of the other numbers are even. Vice versa for when the perfect square is even. odd+even = odd


How III is false:
n= 4
1,2,4
there's 3 of them.

Why are we not considering negative factors what stops us from doing so !!

But for a perfect square such as 36
Factors are 1,2,3,6
Distinct factors = 4 (Even?)

Also is 1 a distinct factor?

you will NEVER use "negative factors" on the gmat.

some number-theory texts consider "negative factors" while others don't. however, the sole authority on the gmat - namely, gmac - does not allow negative factors. period.

it's almost certain that the problem will actually contain the words "positive factors", just to ward off this potential objection, but you should always solve problems with the mindset that ALL factors are positive.



the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. there are nine of them.

here are probably the fastest (not necessarily easiest) ways to address these.


this is true.

fastest way: just memorize this as a takeaway. in fact, it works BOTH ways:
* if a number has an odd # of total factors, then it's a perfect square
* if a number is a perfect square, then it must have an odd # of total factors

second fastest way: pair off all the factors into pairs that multiply to give the original number. since it's a perfect square, the square root will be by itself, while all the other numbers are in pairs.
for instance, if n = 36:
1, 36
2, 18
3, 12
4, 9
6

perfect squares will always have this lone # at the end, creating an odd number of overall factors. non-perfect-squares will not have it, creating an even number of overall factors (since all factors will be paired off).



actually yes, this must be true.

this is hard to prove - the best way to approach the problem is just to throw a bunch of perfect squares at it, and observe that this works every time - but here's a proof:

background fact: if a, b, c, ... are the exponents in a prime factorization, then the number of factors is (a + 1)(b + 1)(c + 1)...
for instance, 36 is (2^2)(3^2), so a = 2 and b = 2. therefore, the total number of factors is (2 + 1)(2 + 1), or 9.
in a perfect square, all of a, b, c, ... are EVEN. this is what makes a number a perfect square: every prime is raised to an even power.
so all of a + 1, b + 1, ... are odd.
so the total number of factors is a product of a bunch of odd numbers, and is therefore itself odd.



this is false.

in a perfect square, you can have any collection of prime factors that you want.

so (c)

hey!!
just wanted to know since statement no II of the question lacks the word "distinct" shouldnt we sum up all the factors including the one that is not distinct..
For example:the sum of factors of 9 should be 1+ 9+3+3=16 (even).
i think Statement II can be true only if the word "distinct" is mentioned there just like in statement I and III.

Please reply.
thanks.

Its very tricky question but answer is really very simple