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Perfect Sq

Expert replies
Source: — Data Sufficiency |

by Osirus@VeritasPrep » Tue Jan 26, 2010 7:48 am
This was a quick one. You can look at Statement (1) and instantly know that it is sufficient. ALL perfect squares have an odd number of factors. Lets use 16 as an example. Use factor pairs

1 16
2 8
4 4

Because of the perfect square, in this case 4, you end up with 4 on both sides of the factor pair leaving you with an odd number of factors. This will be true for all perfect squares. Statement (1) sufficient.

With statement (2) I didn't remember the rule, so I just chose two perfect squares, an even and an odd one and added together their distinct factors. Lets use 4 and 9.

Distinct factors of 4 are 1, 2, and 4. They sum to 7. Which is odd. Lets use 9 now. 9's distinct factors are 1,3, and 9. They sum to 13, which is odd. You can keep testing, but the sum of distinct factors of perfect squares will always be odd. Therefore N is not a perfect square. Statement (2) is sufficient. Choose D
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by ajith » Tue Jan 26, 2010 11:51 am
sumanr84 wrote: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.

All the numbers can be expressed in a way

a^x*b^y*c^z....
a,b,c being prime and x,y,z ...an integer greater than or equal to 1

If the number is a square and x,y,z... all will be even numbers

so, the number of distinct factors will be (x+1)*(y+1)*(z+1).... Will be odd

If the number is not a square one of x,y,z... will be even thus making the num of factors even (x+1)*(y+1)*(z+1)....

So

1 is sufficient to conclude that the number is not a square


--------

as far as 2 goes it says that sum of all distinct factors = even

For all squares sum of the factors = odd

that is because

All the squares can be expressed in a way

a^x*b^y*c^z....
a,b,c being prime and x,y,z ...all are even numbers


now sum of factors= (x^0+x^1+.....x^a) (y^0+y^1+....y^b)......

Now (x^0+x^1+.....x^a) is odd because it is 'a' is even
same goes for (y^0+y^1+....y^b)

hence sum of factors for a perfect square is always odd

So 2 is also sufficient to conclude that the number is not a square

[I know the explanation contains a lot of math, I am extremely sorry and this will most probably not be a GMAT Question]
Always borrow money from a pessimist, he doesn't expect to be paid back.
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