The prime digits are 2, 3, 5, and 7.

The perfect square digits are 0, 1, 4, and 9.

The only digits not represent in these two groups are 6 and 8, which are, coincidentally, found in the number 68.

By observing each of these sets, we can easily determine the largest number in each; however, the problem is asking for us to find the greatest possible sum if we select four different integers, one from each set. The largest number is 11. We will either select this number from set Y or set Z.

We will select set

Y for the 11

because the next larger number in set

Z is 10, which is greater than the next largest number is set Y, 7.

We can use set Z for 10.

Because the 10 has already been selected, choosing the integer from W and X should be easy because 9 is the next largest integer from set W

and 

8 is the next largest integer from set X.

11 + 10 + 9 + 8 = 38

If a team's ratio of wins to losses increases from  to  by winning six games, they must have 12 wins initially and 3 losses. 

The rule is as follows: add 2 to the previous number then multiply times 2.

The answer is (16 + 2) * 2 = 36.

We know the average of Set A is 5. Therefore:

Now let's solve for :

We can now plug this value of  into Set B and solve for the average:

Set B = {0, 2, 4}

The average of Set B is 2.

Three consecutive even integers can be represented by x, x+2, x+4.  The sum is 3x+6, which is equal to 108.  Thus, 3x+6=108.  Solving for x yields x=34.  However, the question asks for the largest number, which is x+4 or 38.  Please make sure to answer what the question asks for!

You could have also plugged in the answer choices.  If you plugged in 38 as the largest number, then the previous even integer would be 36 and the next previous even integer 34.  The sum of 34, 36, and 38 yields 108.

Let us call x the smallest integer. Because the next two numbers are consecutive even integers, we can call represent them as x + 2 and x + 4. We are told the sum of x, x+2, and x+4 is equal to 72. 

x + (x + 2) + (x + 4) = 72

3x + 6 = 72

3x = 66

x = 22.

This means that the integers are 22, 24, and 26. The question asks us for the product of these numbers, which is 22(24)(26) = 13728. 

The answer is 13728.

Four consecutive integers could be represented as n, n+1, n+2, n+3

Therefore, by saying that they have a mean of 9.5, we mean to say:

(n + n+1 + n+2 + n+ 3)/4 = 9.5

(4n + 6)/4 = 9.5 → 4n + 6 = 38 → 4n = 32 → n = 8

Therefore, the largest value is n + 3, or 11.

Let x be the smallest of the four integers. We are told that the integers are consecutive odd integers. Because odd integers are separated by two, each consecutive odd integer is two larger than the one before it. Thus, we can let x + 2 represent the second integer, x + 4 represent the third, and x + 6 represent the fourth. The sum of the four integers equals 96, so we can write the following equation:

x + (x + 2) + (x + 4) + (x + 6) = 96

Combine x terms.

4x + 2 + 4 + 6 = 96

Combine constants on the left side.

4x + 12 = 96

Subtract 12 from both sides.

4x = 84

Divide both sides by 4.

x = 21

This means the smallest integer is 21. The other integers are therefore 23, 25, and 27.

The question asks us how many of the four integers are prime. A prime number is divisible only by itself and one. Among the four integers, only 23 is prime. The number 21 is divisible by 3 and 7; the number 25 is divisible by 5; and 27 is divisible by 3 and 9. Thus, 23 is the only number from the integers that is prime. There is only one prime integer.

The answer is 1. 

Assume the three consecutive integers equal , , and . The sum of these three integers is 60. Thus,