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) 

II) 

III) 

IV) 

Disregard I and IV since each repeats a factor. 

In (II), the sum of the factors is ; in (III) the sum is . 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:

Sum: 

Sum: 

Sum: 

Sum: 

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:

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

.

Therefore, 72 is the greatest common factor.

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

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

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

Thus:

We substitute these equalities into the given expression and simplify.

Since  and  are two-digit integers (equal to  and respectively), we must have  and . Any other factor values for or will produce three-digit integers (or greater).

is equal to , so  could be either 1 or 2. 

Therefore:

or 

The first seven primes are 2, 3, 5, 7, 11, 13, and 17. Don't forget about 2, the smallest prime number, and also the only even prime! Adding these seven numbers gives a sum of 58, and 58/2 = 29.

Plug in a prime number such as  and evaluate all the possible solutions. Note that the question asks which value COULD be prime, not which MUST BE prime. As soon as your number-picking yields a prime number, you have satisfied the "could be prime" standard and know that you have a correct answer.

There are 8 possible numbers; 4,6,8,10,12,14,16,18.

One is not a prime number, so only 8, 10, 12, 14, 16, and 18 can be the sum of two different prime numbers.

The primes, in order, are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, ...

We create a few series:

 -> series length 2: 2,3

 -> series length 3: 3,5,7

 -> series length 5: 5,7,11,13,17

 -> series length 7: 7,11,13,17,19,23,29

etc.

We can then see that, of the answers, only 47 and 31 remain possibly correct answers.  Now we need to decide which of those two are impossible. 

We could do another series, but the  series has 11 terms requiring us to go further and further up.  If we do this, we'll find that it terminates at 47, meaning that 31 must be the correct answer.

Another way, however, is to notice that 29 is the end of the  series.  Since 31 is the very next prime number, if we start on 11, the series that terminates in 31 would have to have length 7 as well.  Every series after  will thus end on a number larger than 31, meaning we will never finish on a 31.