Let's start by writing the prime factorization of these two numbers: 

\(\displaystyle 120=2^{3}\cdot3\cdot5\) and \(\displaystyle 268=2^{2}\cdot67\).

Then we just have to multiply each factor raised to its highet power for common factors, as follows:

\(\displaystyle 2^{3}\cdot3\cdot5\cdot67\) or \(\displaystyle 8040\).

For any positive integer \(\displaystyle a\) and \(\displaystyle b\), \(\displaystyle LCM\) being their least common multiple and \(\displaystyle GCF\) being their greatest common factor, then \(\displaystyle a\cdot b=LCM\cdot GCF\).

Therefore,

 \(\displaystyle LCM=\frac{96}{4}\) or \(\displaystyle 24\).

We can use the property that for \(\displaystyle a\) and \(\displaystyle b\), whose greatest common factor is \(\displaystyle GCF\) and whose least common multiple is \(\displaystyle LCM\), then \(\displaystyle a\cdot b= LCM\cdot GCF\).

Therefore,

 \(\displaystyle LCM=\frac{338}{13}\) or \(\displaystyle 26\).

Using the property

 \(\displaystyle a\cdot b= LCM(a,b)\cdot GCF(a,b)\) then, \(\displaystyle GCD(a,b)= \frac{5832}{648}= 9\).

Notice that the GCD or greatest common divisor is the same thing as the GCF, or greatest common factor.

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

For example, \(\displaystyle 16\) has the following factors, \(\displaystyle 1, 2, 4, 8, 16\).

There are \(\displaystyle 5\) distinct factors, which is an odd number.

The only perfect square in the answer choices is \(\displaystyle 16\) , 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 : \(\displaystyle 5, 10, 15, 20, 25, 30\). 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 \(\displaystyle 7\).

We must remember that the number of zeros at the end of \(\displaystyle 50!\) 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: 

\(\displaystyle 5, 10, 15, 20, 25, 30, 35, 40, 45, 50\). There are 10 multiples of 5, but in reality, since 25 and 50 both have \(\displaystyle 5^{2}\) in their prime factorization, there are 12 fives in the product; therefore, there will be \(\displaystyle 12\) zeros at the end of \(\displaystyle 50!\), 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 \(\displaystyle n\) for which our product will be zero, namely for \(\displaystyle n=1\) and \(\displaystyle n=2\). Remember that \(\displaystyle n\) 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 \(\displaystyle n\). Since the product is the product of two consecutive integers, the result will always be even; in other words it will be divisible by \(\displaystyle 2\). Therefore, the final answer is \(\displaystyle 2\).

We don't know \(\displaystyle n\), 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, \(\displaystyle 6\) is the final answer.

If \(\displaystyle a+b\) is even, then either both \(\displaystyle a\) and \(\displaystyle b\) are even or both are odd. 

If both \(\displaystyle a\) and \(\displaystyle b\) are even, then: 

(a) \(\displaystyle a-1\) and \(\displaystyle b+ 1\) are odd, and their product \(\displaystyle (a-1)(b+1)\) is odd.

(b) \(\displaystyle ab\) is even, both \(\displaystyle ab-1\) and \(\displaystyle ab+1\) are odd, and their product \(\displaystyle (ab-1)(ab+1)\) is odd.

(c) \(\displaystyle a^{2}\) and \(\displaystyle b^{2}\) are even, \(\displaystyle a^{2}-1\) and \(\displaystyle b^{2}+ 1\) are odd, and their product \(\displaystyle (a^{2}-1)(b^{2}+1)\) is odd.

If both \(\displaystyle a\) and \(\displaystyle b\) are odd, then: 

(a) \(\displaystyle a-1\) and \(\displaystyle b+ 1\) are even, and their product \(\displaystyle (a-1)(b+1)\) is even.

(b) \(\displaystyle ab\) is odd, both \(\displaystyle ab-1\) and \(\displaystyle ab+1\) are even, and their product \(\displaystyle (ab-1)(ab+1)\) is even.

(c) \(\displaystyle a^{2}\) and \(\displaystyle b^{2}\) are odd, \(\displaystyle a^{2}-1\) and \(\displaystyle b^{2}+ 1\) are even, and their product \(\displaystyle (a^{2}-1)(b^{2}+1)\) is even.

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