This will require us to know the divisibility rule of . The reason for this choice is that some of the numbers are palindromes like  so we eliminate . For the three digit numbers, the divisibility rule for  is add the outside digits and if the sum matches the sum then it is divisible. Let's see.

Based on this test,  is not divisible by  and is our answer.

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.

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. 

Let x be the first integer. The next two integers would then be x+1 and x+2. We must find x such that:

x(x+1)(x+2) = 210.

To do this, we need to look at the factors of 210, and find a set of three factors that are also consecutive integers. First, let's find the prime factorization of 210, because then we can use that to determine all of the possible factors of 210.

210 = 10 x 21 = 2 x 5 x 21 = 2 x 5 x 3 x 7

Thus, the prime factorization of 210 is 2 x 3 x 5 x 7.

We can notice now that 5 and 7 are almost consecutive integers, and the only number missing between them is 6. If we multiply 2 and 3, we get 6. Thus 210 can be written as 5 x 6 x 7. This means that the consecutive integers are 5, 6, and 7, and their sum is 18.

For this question, use the fact that a sum of two multiples is a multiple. In other words:

if x is a multiple of 3 and y is a multiple of 3: (x + y) is a multiple of 3.

Thus in this question, x is a multiple of 2 and 3. We need to find a number that is a multiple of 2, 3, and 5. 

Take 5x + 30:

[x is divisible by 2. 5 times x is still divisble by 3. 30 is divisible by 2.] -> divisible by 2.

[x is divisible by 3. 5 times x is still divisble by 3. 30 is divisible by 3.] -> divisible by 3.

[5x is divisible by 5. 30 is divisible by 5.] -> divisible by 5.

Consider the possible values for (x, y):

(1, 100)

(2, 50)

(4, 25)

(5, 20)

Note that (10, 10) is not possible since the two variables are to be distinct. The sums of the above pairs, respectively, are:

1 + 100 = 101

2 + 50 = 52

4 + 25 = 29

5 + 20 = 25, the smallest possible value.

Let n represent all of the numbers that are equal to (6 + 2p)(3). Then, let's solve for p in terms of n.

(6 + 2p)(3) = n

Divide both sides by 3.

(6 + 2p) = n/3

Subtract six from both sides.

2p = –6 + n/3

Divide both sides by 2.

p = –3 + n/6

Since we are told that p is an integer, the only way that p can be an integer is if n is a multiple of 6. Only a multiple of six, when divided by six, will yield an integer number. Any integer plus –3 will also be an integer, since the sum of two integers is always an integer.

In short, we must look for the answer choice that is a multiple of 6. Of the choices, only 84 is a multiple of 6.

The answer is 84.