Another property of perfect squares, is that they will always have an odd number of distinct factors.

For example,  has the following factors, .

There are  distinct factors, which is an odd number.

The only perfect square in the answer choices is  , therefore, it is the final answer.

Firstly, we must remember that the number of zeros at the end will be given by the number of powers of ten that we have in the factorial of 30. Ten is the product of 2 and 5, and since there will be more powers of two in the product than powers of 5, the final answer is given by the number of powers of 5 in the product. (This is because a two will then multiply each of the fives, which would give us 10 raised to the power of however many five we have.) We can write them out : . Now we would be tempted to answer 6, since there are 6 multiples of 5, but in fact there are 7 fives since 25 is the product of two fives. 

Therefore, the final answer is .

We must remember that the number of zeros at the end of  will be given by the number of powers of ten that we have in the factorial of 50. Ten is the product of 2 and 5, and since there will be more powers of two in the product than powers of 5, the final answer is given by the number of powers of 5 in the product. (This is because a two will then multiply each of the fives, which would give us 10 raised to the power of however many fives we have.)

We should write out all the multiples of 5 in the product as follows: 

. There are 10 multiples of 5, but in reality, since 25 and 50 both have  in their prime factorization, there are 12 fives in the product; therefore, there will be  zeros at the end of , since 12 even numbers will then multiply our 12 fives, which would give us 10 raised to the power of 12.

Here, the tricky part is that we might get confused by small values for  for which our product will be zero, namely for  and . Remember that  can't be zero since zero is not a positive number. But actually, zero can be divided by any number. Therefore, we must take a look at greater values for . Since the product is the product of two consecutive integers, the result will always be even; in other words it will be divisible by . Therefore, the final answer is .

We don't know , but we can see that it is the product of three consecutive integers; therefore, the product will either include two odd numbers and one even number or two even numbers and one odd number. This product can be divided by 3 and 2 or 6, since from three consecutive integers, at least one will be 3. Therefore,  is the final answer.

If  is even, then either both  and  are even or both are odd. 

If both  and  are even, then: 

(a)  and  are odd, and their product  is odd.

(b)  is even, both  and  are odd, and their product  is odd.

(c)  and  are even,  and  are odd, and their product  is odd.

If both  and  are odd, then: 

(a)  and  are even, and their product  is even.

(b)  is odd, both  and  are even, and their product  is even.

(c)  and  are odd,  and  are even, and their product  is even.

Therefore, all three expressions may be even or odd, and the correct response is none.

We recognize the expression for  as a perfect square trinomial which can be rewritten as:

 is odd if and only if the square root  is odd. This happens if and only if one of  and  is even and one is odd - this is the correct response.

If  and  are both odd, then , the product of odd numbers, is odd, and , the sum of three odd numbers, is odd.

If  and  are both even, then , the product of even numbers, is even, and , the sum of three even numbers, is even.

If one of  and  is even and one is odd, then , which has an even factor, is even, and , the sum of two even numbers and an odd number, is odd.

Therefore,  is even if and only if  and  are both even.

For the product of two positive integers to be odd, both of the integers must themelves be odd. Therefore, if is odd, it follows that both and are odd as well. We examine each of the quantities keeping this in mind.

(a) Both and are odd, so  and  are even. Their product, which is , must be even.

(b) Since  is odd, both  and  are even. Their product,   , is even.

(c) Since both and are odd, their squares and are both odd; it follows that  and  are both even, so their product is even.

The correct response is that all of the three quantites must be even.

is even, so it follows that one of and is even; the other can be even or odd. Let us assume that will always be even; the argument will be similar if is assumed to always be even.

(a)  is even, so is odd. If is even, then  is odd, and , the product of odd factors, is odd. If is odd, then  is even, and , having an even factor, is even. Therefore, can be even or odd.

(b) is even, so and are both odd. is the product of odd factors and must be odd.

(c) is even, so is even as well, and is odd. If is even, then  is even, and   is odd; , the product of odd factors, is odd. If is odd, then  is odd, and  is even; , having an even factor, is even. Therefore,  can be even or odd.

The correct response is that only (b) need be odd.