# Find Four Consecutive Even Integers Whose Sum is 8012 by 3 Ways

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The sum of four consecutive even integers is 8012, what are these even integers? I can easily tell you that the answer are 2000, 2002, 2004 and 2006. You must be interested in how to find these 4 consecutive even integers whose sum is 8012. There are three methods here, let us introduce them one by one below.

## How to find four consecutive even integers whose sum is 8012?

### 1. Hypothetical method

Assuming that 2 * N is used to represent the first even integer, so the second even integer is 2 * N + 2, the third even integer is 2 * N + 4, the 4th even integer is 2 * N + 6.

Therefore, the sum of 4 consecutive even integers is 8012, which can be expressed by the equation

2 * N + (2 * N + 2) + (2 * N + 4) + (2 * N + 6) = 8012

Solve this equation

2 * N + (2 * N + 2) + (2 * N + 4) + (2 * N + 6) = 8012

2 * N + 2 * N + 2 + 2 * N + 4 + 2 * N + 6 = 8012

4 * 2 * N + 2 + 4 + 6 = 8012

8 * N + 12 = 8012

8 * N = 8012 – 12

8 * N = 8000

N = 8000 / 8

N = 1000

So, 2 * N = 2 * 1000 = 2000

Now, we can get that 2000 is the first integer of 4 consecutive even integers whose sum is 8012. So, the second even integer is 2002, the third even integer is 2004, the 4th even integer is 2006.

The answer came out, the sum of 4 consecutive even integers is 8012, these three even integers are 2000, 2002, 2004 and 2006.

Verify: 2000 + 2002 + 2004 + 2006 = 8012. Correct!

This is the most common method, let’s look at the second method, which is my favorite method.

### 2. Formula method

According to the consecutive integers calculator based on sum, we can know that the sum of M consecutive even integers is S, the first even integer formula is

First(e) = S / M – M + 1

M represents the number of consecutive even integers.

S stands for sum.

Back to the problem we want to solve: find four consecutive even integers whose sum is 8012. In here, M = 4, S = 8012. Replace them in the formula to calculate the first even integer

First(e) = 8012 / 4 – 4 + 1 = 2003 – 3 = 2000

The first even integer is 2000, so, it can be easily calculated that the second even integer is 2002, the third even integer is 2004, the 4th even integer is 2006.

Thus, the answer are also 2000, 2002, 2004 and 2006. Same as the first method. Then look at the third method, which is the simplest one.

### 3. Average method

The principle is very simple, because we want to calculate the consecutive even integers, so after calculating the average, we can find the even integers near the average.

The sum of 4 consecutive even integers is 8012, so, the average of these 4 consecutive even integers is 8012 / 4 = 2003.0. The even integers around 2003 are 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009. Now the problem is simple, find 4 consecutive even integers from 1997 to 2009 and their average is 2003. The answer are 2000, 2002, 2004 and 2006.

So the sum of 4 consecutive even integers is 8012, these even integers are 2000, 2002, 2004, 2006. The results are consistent with the above two methods. Is it very simple?

## Problems can be sloved by this answer

Now, we have found out these 4 consecutive even integers whose sum is 8012, some problems can be easily solved.

• 1. What are four consecutive even integers whose sum is 8012?
The answer are 2000, 2002, 2004 and 2006.
• 2. What is the smallest of four consecutive even integers whose sum is 8012?
The smallest number is 2000.
• 3. What is the greatest of four consecutive even integers whose sum is 8012?
The greatest number is 2006.
• 4. What is the average of four consecutive even integers whose sum is 8012?
The average of these 4 consecutive even integers is (2000 + 2002 + 2004 + 2006) / 4 = 8012 / 4 = 2003.
• 5. What is the product of four consecutive even integers whose sum is 8012?
The sum of 4 consecutive even integers is 8012, the product of them is 2000 * 2002 * 2004 * 2006 = -1361329408.
• 6. What is the sum of square of four consecutive even integers whose sum is 8012?
The sum of 4 consecutive even integers is 8012, the sum of their squares is 20002 + 20022 + 20042 + 20062 = 16048056.

## Summarize

On this page, In addition to introducing three methods how to find four consecutive even integers whose sum is 8012, it also provides a calculator that calculates four consecutive even integers based on the sum. If you encounter a similar problem next time, you can directly use this calculator to calculate the answer, which is very convenient.

Of course, if the problem you encounter is more complicated, such as: the number of consecutive integers is not 4, or you need to calculate consecutive integers or consecutive odd integers. You can use our other more advanced sum-based consecutive integers calculator, where you can specify the number of consecutive integers and select consecutive integers type: natural integers, odd integers or even integers. I believe it can help you.

Well, that’s it, the above are three solutions to find four consecutive even integers whose sum is 8012. Personally, I prefer the second method, because it can accurately calculate the first integer and the calculation process is very simple. So how about you? Please leave a message and tell me which method you like?