The prime factorization of 24 is (2)(2)(2)(3). The distinct primes are 2 and 3, the product of which is 6.

The prime factorization of 24 is (2)(2)(2)(3). The distinct primes are 2 and 3, the product of which is 6.

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.

Thus 5x + 30 is divisible by 2, 3 and 5.

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.

Thus 5x + 30 is divisible by 2, 3 and 5.

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.

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.

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.

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.

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.

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.