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

$\displaystyle 33 = 3 \times 11$

$\displaystyle 35 = 5 \times 7$

$\displaystyle 39 = 3 \times 13$

However, 37, being a prime, has only one factorization - $\displaystyle 1 \times 37= 37$, and 1 is not a prime. 37 is the correct choice.

For example, If $\displaystyle x=6$, the only way to multiply this to another number $\displaystyle y$ and get $\displaystyle 1$ is if $\displaystyle y =\frac{1}{6}$, but $\displaystyle \frac{1}{6}$ is not an integer, so this is impossible. (integers are the counting numbers $\displaystyle 1, 2, 3, 4, ....$, their negatives, and $\displaystyle 0$)

The quotient $\displaystyle \frac{x}{y}$ 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 $\displaystyle A$ be the missing common digit, then the digit sum is

$\displaystyle 7+A +5+A+ 6 + A = 18 + 3 A$

$\displaystyle 18 + 3 A$ divided by 3 is $\displaystyle \frac{18 + 3 A }{3} = 6+A$, 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:

$\displaystyle 705,060 \div 8 = 88,132 \textup{ R }4$

$\displaystyle 745,464 \div 8 = 93, 183$

$\displaystyle 785,868 \div 8 =98,333\textup{ R }4$

Only 4 works, so the correct choice is one.

An even number can be written as $\displaystyle 2m$ where $\displaystyle m$ is an integer. The expression $\displaystyle 2m+6$ is even since it is the sum of two even numbers ($\displaystyle 2m$ and $\displaystyle 6$) and also can be written as $\displaystyle 2(m+3)$. 

$\displaystyle 2m+6$ must be even if $\displaystyle m$ is an integer.

Let's look at the other answers:

$\displaystyle 2m+1$ is odd whether $\displaystyle m$ is odd or even. Therefore, it is not true that $\displaystyle 2m+1$ must be even.

$\displaystyle m^2$ is even only if $\displaystyle m$ is even and odd if $\displaystyle m$ is odd. Therefore, it is not true that $\displaystyle m^2$ must be even.

$\displaystyle m+1$ is even if $\displaystyle m$ is odd and odd if $\displaystyle m$ is even. Therefore, it is not true that $\displaystyle m+1$ must be even.

$\displaystyle 3m$ is even if $\displaystyle m$ is even and odd is $\displaystyle m$ is odd. Therefore, it is not true that $\displaystyle 3m$ must be even.

We know that the remainder of $\displaystyle m$ divided by $\displaystyle n$ is 7.

So we can write $\displaystyle m=nq+7$ , where $\displaystyle q$ is the quotient of $\displaystyle m$ divided by $\displaystyle n$.

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

$\displaystyle 2n \leq m< 3n$

This expression reveals that the quotient of $\displaystyle m$ divided by $\displaystyle n$ is 2 since $\displaystyle m$ is greater than $\displaystyle 2n$ but less than $\displaystyle 3n$.

Therefore:

$\displaystyle m=2n+7$

The last piece of information is that $\displaystyle m-n=16$, So, $\displaystyle m=n+16$.

Replacing $\displaystyle m$ in the previous equation gives:

$\displaystyle n+16=2n+7$

$\displaystyle n+7=16$

$\displaystyle n=9$

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

$\displaystyle x=6n$ and $\displaystyle 15< x< 50$

First, find an odd integer $\displaystyle n$ such that $\displaystyle 15< 6n< 50$ and $\displaystyle n>6$.

(1) Try $\displaystyle n=7$

$\displaystyle x=6\times7=42$

If $\displaystyle n$ is 7, then $\displaystyle x$ is 42 and $\displaystyle 15< x< 50$.

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

(2) Try $\displaystyle n=9$

$\displaystyle x=6\times9=54$

If $\displaystyle n$ is 9, then $\displaystyle x>50$ 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.

$\displaystyle 15 = 3 \times 5$

$\displaystyle 35 = 5 \times 7$

$\displaystyle 55 = 5 \times 11$

$\displaystyle 95 = 5 \times 19$

$\displaystyle 15$, $\displaystyle 35$, $\displaystyle 55$, and $\displaystyle 95$ can all be expressed as the product of two primes. 

However, $\displaystyle 75$ can be factored as the product of two integers in the following ways:

$\displaystyle 75 = 1 \times 75 = 3 \times 25 = 5 \times 15$

In each factorization, at least one number is composite ($\displaystyle 75$, $\displaystyle 25$, $\displaystyle 15$). $\displaystyle 75$ is the correct choice.

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

$\displaystyle 72 = 5+ 67$

$\displaystyle 74 = 3+ 71$

$\displaystyle 76 = 5+71$

$\displaystyle 78 = 7+71$