Since v is divisible by 2, 3 and 15, v must be a multiple of 30. Any number that is divisible by both 2 and 15 must be divisible by their product, 30, since this is the least common multiple.

Out of all the answer choices, v + 30 is the only one that equals a multiple of 30.

We are given information that m/4 is 10 greater than m/3. We set up an equation where m/4 = m/3 + 10.

We must then give the m variables a common denominator in order to solve for m. Since 3 * 4 = 12, we can use 12 as our denominator for both m variables.

m/4 = m/3 + 10 (Multiply m/4 by 3 in the numerator and denominator.)

3m/12 = m/3 + 10 (Multiply m/3 by 4 in the numerator and denominator.)

3m/12 = 4m/12 + 10 (Subtract 4m/12 on both sides.)

-m/12 = 10 (Multiply both sides by -12.)

m = -120

-120/4 = -30 and -120/3 = -40. -30 is 10 greater than -40.

We need to find ways to factor 18 such that the three factors are different, and then find the sum of those factors in each case.

18 can be factored as the product of three integers in four ways:

I) $\displaystyle 1 \times 1 \times 18$

II) $\displaystyle 1 \times 2 \times 9$

III) $\displaystyle 1 \times 3 \times 6$

IV) $\displaystyle 2 \times 3 \times 3$

Disregard I and IV since each repeats a factor. 

In (II), the sum of the factors is $\displaystyle 1 + 2 + 9= 12$; in (III) the sum is $\displaystyle 1 + 3 + 6 = 10$. Of the five choices, only 12 is possible.

We look for ways to write 45 as the product of three integers, then we find the sum of the integers in each situation. They are:

$\displaystyle 1 \times 1 \times 45$

Sum: $\displaystyle 1 + 1 + 45 = 47$

$\displaystyle 1 \times 3 \times 15$

Sum: $\displaystyle 1 + 3 + 15 = 19$

$\displaystyle 1 \times 5 \times 9$

Sum: $\displaystyle 1 + 5 + 9 = 15$

$\displaystyle 3 \times 3 \times 5$

Sum: $\displaystyle 3 + 3 + 5 = 11$

Of the five choices, only 33 is not a  possible sum of the factors. This is the correct choice.

The greatest common factor is the largest factor that both numbers share. Each number has many factors. The factors for 72 are as follows:

$\displaystyle 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$

Starting from the largest factor, 72, we can see that it is also a factor of 144

$\displaystyle 144 = (72 \times 2)$.

Therefore, 72 is the greatest common factor.

The greatest common divisor of $\displaystyle x$ and $\displaystyle y$ is 10. This means that the prime factorizations of $\displaystyle x$ and $\displaystyle y$ must both contain a 2 and a 5. 

The greatest common divisor of $\displaystyle y$ and $\displaystyle z$ is 9. This means that the prime factorizations of $\displaystyle y$ and $\displaystyle z$ must both contain two 3's.

The greatest common divisor of $\displaystyle x$ and $\displaystyle z$ is 8. This means that the prime factorizations of $\displaystyle x$ and $\displaystyle z$ must both contain three 2's.

Thus:

$\displaystyle x=2^3*5*A,\ y=2*3^2*5*B,\ z=2^3*3^2*C$

We substitute these equalities into the given expression and simplify.

$\displaystyle \frac{yz}{x}=\frac{(2*3^2*5*B)(2^3*3^2*C)}{(2^3*5*A)} = \frac{162BC}{A}$

Since $\displaystyle y$ and $\displaystyle z$ are two-digit integers (equal to $\displaystyle 90B$ and $\displaystyle 72C$respectively), we must have $\displaystyle B=1$ and $\displaystyle C=1$. Any other factor values for $\displaystyle B$ or $\displaystyle C$ will produce three-digit integers (or greater).

$\displaystyle x$ is equal to $\displaystyle 40A$, so $\displaystyle A$ could be either 1 or 2. 

Therefore:

$\displaystyle \frac{yz}{x}=162$

or 

$\displaystyle \frac{yz}{x}=81$