Number of non-square factors?

Nice one, this. You need to do 2 things, find the prime number factors, and find the square root. As far as the prime number factors are concerned, there are 8 2s, 2 3s and 6 5s. The virtue of knowing the square root is that if there is a factor less than the square root, there is a matching factor greater than the square root, so you only have half the work.

Now you select combinations of the prime factors which contain only odd numbers of each factor, which by definition will not be perfect squares, e.g.

2

2 x 2 x 2

2 x 2 x 3

2 x 3

are not perfect squares, and neither will the matching factors above the square root be perfect squares, whereas

2 x 2

2 x 2 x 3 x3

are perfect squares.

As a final thought, prove that a perfect square always has an odd number of factors. Answer: the perfect square n has at least one factor below its square root, which is 1, and one above, which is n. All other factors are in pairs, one above and one below, with the exception of the square root, which is single.

Let x = The number Ten times the square..: 10 * ( x^2) is equal to.. 10 * (x^2) = Nintey times the number 10 * (x^2) = 90x 10x^2 = 90x 10x^2 - 90x = 0 10x(x - 9) = 0 10x = 0 ; (x - 9) = 0 x = 0 ; x = 9 Since x is a "non-zero" number, x is 9 x = 9

36000000 = 2^6 * 5^6 * 6^2

Just find all possible products of these primes where at least one of their exponents is odd. I won't bother to write them all out for you.