bblast wrote:Cant recall a concept :

A perfect square has odd number of factors,

Is the opposite also true ? Any integer with odd number of factors is a perfect square ? I think yes.

This has the ring of truth (at least for positive integers). Generally speaking, factors come in "pairs" of two integers, the product of which reaches the original number.

For example,
6 can be split into two pairs of factors: 2 and 3, 1 and 6.
48 can be split into five pairs of factors: 1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8.

The only way an integer will have an odd number of factors is if one of the pairs is actually the same factor times itself: 36 splits into 1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6. - which would make it a perfect square.

thank you Geva. You actually pointed the root of the theory behind odd no of factors.

So if i devise a DS question :

If x is a positive integer, is x a perfect square ?

1>x has odd no of factors.
2>no of factors of x = 3.


statement 1 is sufficient, as per discussion.
fancy statement 2 means x = square of a prime and hence, sufficient. sorry if the question is really lame.

bblast wrote:thank you Geva. You actually pointed the root of the theory behind odd no of factors.

So if i devise a DS question :

If x is a positive integer, is x a perfect square ?

1>x has odd no of factors.
2>no of factors of x = 3.


statement 1 is sufficient, as per discussion.
fancy statement 2 means x = square of a prime and hence, sufficient. sorry if the question is really lame.

Wouldn't dream of making fun even if I did think it was lame, but it's actually a good question testing a solid concept. Stat. (2) is a specific case of stat. (1) because 3 is odd; meaning that if we accept that (1) is sufficient, then (2) is also sufficient by default.

statement 2 would have been slightly better if it read

x is the square of a prime number. - in this way we are not directly stating an odd no of factors.

Anyways, the topic is clear. Thank you sir.