Three of the choices can be written as the product of two distinct primes:

However, 37, being a prime, has only one factorization - , and 1 is not a prime. 37 is the correct choice.

For example, If , the only way to multiply this to another number and get is if , but is not an integer, so this is impossible. (integers are the counting numbers , their negatives, and )

The quotient  is negative if x and y have opposite signs; that is if x is positive, y will be negative or if x is negative y will be positive.

In which case, the product of x and y must be negative, that is xy<0.

So the product of x and y cannot be positive in this case.

The other inequalities do not have to be true for the quotient of x and y to be negative.

An integer is divisible by 6 if and only if it is divisible by 2 and 3.

It is divisible by 3 if and only if the sum of its digits is divisible by 3. If we let  be the missing common digit, then the digit sum is

 divided by 3 is , an integer, so the digit sum - and the number itself - will be divisible by 3 regardless of the digit that is entered in all three circles. 

The number will be divisible by 2 if and only if the final digit - thiat is, the digit entered in all three circles - is 0, 2, 4, 6, or 8. 

Therefore, there are five possible digits that can be entered into all three circles to yield a multiple of 6.

For an integer to be divisible by 8, it must also be divisible by 2 and by 4.

The number is a multple of 2, so the last digit - and the common digit - can be narrowed down to 0, 2, 4, 6, and 8.

The number is a multple of 4, so the last two digits must form a multiple of 4; since 60, 64, and 68 are divisible by 4, and 62 and 66 are not, this narrows the choice to 0, 4, and 8.

We can now just try all three cases with straightforward division:

Only 4 works, so the correct choice is one.

An even number can be written as  where  is an integer. The expression  is even since it is the sum of two even numbers ( and ) and also can be written as . 

 must be even if  is an integer.

Let's look at the other answers:

 is odd whether  is odd or even. Therefore, it is not true that  must be even.

 is even only if  is even and odd if  is odd. Therefore, it is not true that  must be even.

 is even if  is odd and odd if  is even. Therefore, it is not true that  must be even.

 is even if  is even and odd is  is odd. Therefore, it is not true that  must be even.

We know that the remainder of  divided by  is 7.

So we can write  , where  is the quotient of  divided by .

The next piece of information tells us that  is at least twice the value of  but less than three times the value of . We can then write the following expression:

This expression reveals that the quotient of  divided by  is 2 since  is greater than  but less than .

Therefore:

The last piece of information is that , So, .

Replacing  in the previous equation gives:

Let  be the number of students in the class. The students can be divided into 6 groups of  members, with  being an odd integer greater than 6.

 and

First, find an odd integer  such that  and .

(1) Try 

If  is 7, then  is 42 and .

If each group has 7 members, there are 42 students in the class.

(2) Try 

If  is 9, then  and does not satisfy the condition of being between 15 and 50.

Therefore, the students can be divided in 6 groups of 7 members. There are 42 students in the class.

, , , and  can all be expressed as the product of two primes. 

However,  can be factored as the product of two integers in the following ways:

In each factorization, at least one number is composite (, , ).  is the correct choice.

Each of the four numbers can be expressed as the sum of two primes. For example: