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.

The value of , or , is the product of  and , so it will be divisible by 1, p, p * p, and nothing else (we know that the p’s are not divisible because they are prime). Therefore p² has exactly three factors.

(Alternatively, we can plug in any prime number for p and see how many factors p² has. For example, if p is 3, then the factors of p², or 9, are 1, 3, and 9.)

Let x represent the smallest of the four numbers. 

Then we can set up the following equation:

Therefore the four numbers are 51, 52, 53, 54.  The only prime in this list is 53.

It is important to know what a prime number is in order to answer this question. A prime number is defined as any positive integer that is divisible only by the number 1 and itself. For example, 17 is a prime number because its only factors are 1 and 17.

The first five prime numbers are 2, 3, 5, 7, and 11. Remember, 1 is not a prime number, and 2 is the only even prime.

Therefore, the sum of the five smallest prime numbers is 28.

Find the prime factors of 42: 7, 3, 2

Find the prime factors of 55: 5, 11

Product of the greatest factors: 7 and 11 = 77

3 is a prime number that is easy to work with, so we can plug that in for p. p⁴ = 81. The positive factors of 81 are 1, 3, 9, 27, and 81. Thus, the answer is five factors.

Let's look at another prime number to plug in for p. If we plug in 2, another easy prime number to work woth, we get p⁴ = 16. The positive factors of 16 are 1, 2, 4, 8 and 16.

Notice that when p = prime number greater than 1, the positive factors for p⁴ are 1, p, p², p³ and p⁴. Multiplying pⁿ by p only adds pⁿ as a factor when p is prime, so you will have n factors plus 1 which is a factor, so p⁴ has 5 factors.

Since 152 is divisible by 2, start by dividing 152 by 2 which gives you factors of 2 and 76. 2 is a prime factor (cannot be divisible by anything other than itself and 1) but 76 can still be divided by 2. Continue dividing until there are only prime factors. You should get prime factors of 2, 2, 2, and 19. 2+2+2+19 = 25.

If y = 2, then x³z = 250. This means that x could be equal to 1, and z could be equal to 250. This would make x, y, and z distinct positive integers. Thus, it is possible for y to equal 2.

If y = 5, then x³z = 40. This means x could be 1, and z could be 40. This would make x, y, and z distinct positive integers.  Therefore, y can equal 5.

If y = 10, then x³z = 10. This means that the only value that x could be is 1. If x were equal to 2, for example, then z would be equal to 10/8, which is not an integer. The same would be true if x were equal to anything other than 1. Thus, x must equal 1, and z must equal 10. However, because y and z must be distinct, y cannot equal 10. 

Thus, y could equal only 2 and 5.