Recall that an integer $d$ is said to be a divisor of an integer $a$ if $a/d$ is also an integer. For how many integers $a$ between $-200$ and $-1$ inclusive is the product of the divisors of $a$ negative?

I have no idea how to approach this problem. Any helps is greatly appreciated.


I think you only have to check how many divisors it has.

If there is an even number of divisors, then the product of an even number of negatives is positive. The product of an odd number of negatives is negative.

The only numbers that have an odd number of divisors are perfect squares, so it suffices to find all the negative perfect squares

The largest one is then $- (14)^2 = -196$, which has factors $-1, -2, -4, -7, -14, -28, -49,-98, -196 $ ($9$ numbers)

Therefore, there are $14$ integers.


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