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....
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Perfect square
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DanaJ
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As someone has outlined in a previous thread, only perfect squares have an odd number of factors, since the square root can only be counted once.
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
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
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cramya
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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
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to the posters above:
yes.
--
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.
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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.
--
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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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
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sureshbala
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Folks, perfect squares will always have odd number of factors and the converse is also true.
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.
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.
