n is a positive integer and also a perfect square, is the factors of n a:

a) odd

b) square number

c) prime

OA A but as far as I know, 1 is also a perfect square and 1 only has 2 factors....

## Perfect square

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- DanaJ
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Let's take an example:

4 has the following factors: 1, 2 and 4

6: 1, 2, 3, 6

9: 1, 3, 9

10: 1, 2, 5, 10

1 has only one factor which is itself which makes it odd1 is also a perfect square and 1 only has 2 factors....

Like Dana points out all perfect squares have odd number of factors. Sometimes this property will come in handy in DS probs also.

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- lunarpower
- GMAT Instructor
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to the posters above:

yes.

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1 only has one factor, namely, 1.

i'm trying to figure out what are the "two factors" to which you referred in your post. if those two factors are 1 and -1, keep in mind that

as far as i've seen, the problem will always state this condition explicitly (i.e., it will say "positive factors", not just "factors")

if you weren't thinking +/- 1, then i'm curious as to what you thought was the second factor.

--

also:

the "perfect squares have odd numbers of factors" fact is obscure, but it's a very nice thing to know. while it's not particularly likely that you'll get to apply this fact, it can make a hard problem quite easy, if you're lucky enough to see a problem on which you can employ it.

the best thing is that it works both ways; i.e.:

* if an integer has an odd # of factors, then it's a perfect square, AND

* if an integer is a perfect square, then it has an odd # of factors.

yes.

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1 only has one factor, namely, 1.

i'm trying to figure out what are the "two factors" to which you referred in your post. if those two factors are 1 and -1, keep in mind that

**gmat problems deal exclusively with POSITIVE "factors"**.as far as i've seen, the problem will always state this condition explicitly (i.e., it will say "positive factors", not just "factors")

if you weren't thinking +/- 1, then i'm curious as to what you thought was the second factor.

--

also:

the "perfect squares have odd numbers of factors" fact is obscure, but it's a very nice thing to know. while it's not particularly likely that you'll get to apply this fact, it can make a hard problem quite easy, if you're lucky enough to see a problem on which you can employ it.

the best thing is that it works both ways; i.e.:

* if an integer has an odd # of factors, then it's a perfect square, AND

* if an integer is a perfect square, then it has an odd # of factors.

Ron has been teaching various standardized tests for 20 years.

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Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

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*Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.*Yves Saint-Laurent

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- sureshbala
- Master | Next Rank: 500 Posts
**Posts:**319**Joined:**04 Feb 2009**Location:**Delhi**Thanked**: 84 times**Followed by:**9 members

Let us consider the prime factorization of a number N i.e

N = p1^a1 x p2^a2 x p3^a3 x ....... x pn^an

The total number of factors of N = (a1+1).(a2+1).(a3+1)........(an+1)

Now if N is a perfect square obviously a1, a2, a3, .....,an all of them must be even.

So (a1+1) , (a2+1), (a3+1),.......(an+1) will be odd and hence their product is odd.

Thus the total number of factors of N is odd.

You can also observe that converse is also true.