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,

There are 5 numbers in the sequnce.

How many numbers are left over if you divide 5 into 457?

There would be 2 numbers!

The second number in the sequence is 9,5,6,2,1

Since  are consecutive integers, we know that at least 2 of them will be even. Since we have 2 that are going to be even, we know that when we divide the product by 2 we will still have an even number.  Since 2 is the only prime that is even, we must have:

What we notice, however, is that for , we have the product is 0.  For , we have the product is 24.  We will then never have a product of 4, meaning that is never going to be a prime number.

Consecutive odd integers can be represented as x, x+2, x+4, and x+6.

We know that the sum of these integers is 32. We can add the terms together and set it equal to 32:

x + (x+2) + (x+4) + (x+6) = 32

4x + 12 = 32

4x = 20

x = 5; x+2=7; x+4 = 9; x+6 = 11

Our integers are 5, 7, 9, and 11.

The difference between each number in the series is 9. You can substract nine 11 times from 100 to get 1: 100 – 9x11 = 1. Counting 100, there are 12 positive numbers in the series.