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by diebeatsthegmat » Tue May 17, 2011 12:15 am
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.

i dont understand statement 2.
stement 2 says all distinct factors of N is even . for exam if n=25 and its factor is 1,5,25 so the sum of those number is odd but if xis even for example: n=4, its sum of all factors also odd
can you guy please explain me why the sum of factors of n ( even its odd or even) is even? i dont understand why the sum of all distinct factors of n could ve even?
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by vineeshp » Tue May 17, 2011 3:50 am
You are actually answering the question.

For a perfect square you yourself have explained that you are always getting odd. So that proves that statement 2 is enough to say whether N is even or not even. Hence statement 2 is sufficient!

(if u take n=3.Then thesum of distinct factors n is 4, even. So as per the statement,it helps us identify whether n is even or not.)
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by sanju09 » Tue May 17, 2011 4:06 am
diebeatsthegmat 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.

i dont understand statement 2.
stement 2 says all distinct factors of N is even . for exam if n=25 and its factor is 1,5,25 so the sum of those number is odd but if xis even for example: n=4, its sum of all factors also odd
can you guy please explain me why the sum of factors of n ( even its odd or even) is even? i dont understand why the sum of all distinct factors of n could ve even?

(1) All the perfect squares can be written in the form p^a × q^b × r^c × ... where p, q' r ... are distinct primes and a, b, c ... are even positive integers and the number of distinct factors of such a number can be given by (a + 1) (b + 1) (c + 1) ..., which is odd × odd × odd × ... = an odd.

If N has got an even number of distinct factors, then N is NOT a perfect square as per the above argument. Sufficient

(2) if a perfect square is odd, then it has got an odd number of odd factors whose sum is always an odd. And if the perfect square is even, then it has got an even number of even factors and hence an odd number of odd factors in order to maintain the fact that a perfect square has got an odd number of distinct factors. This way, the sum of all factors of an even perfect square is even × even + odd × odd = even + odd = an odd.

As a result, sum of all distinct factors of a perfect square is always odd; and if the sum of all distinct factors of N is even, then N is NOT a perfect square as per the above argument. [spoiler]Sufficient


D[/spoiler]
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