show that the sum of four consecutive whole numbers is divisible by 2

Kendra,

piggy-backing on other students' questions has two disadvantages:
1. Tutors do not get to see your post (as in the present case)
2. it slows down the other student's getting a response.

Your other post has been answered here:
http://www.jiskha.com/display.cgi?id=1322609635

Why did the mathematician get a puppy? Because he wanted a friend to sum it up with!

Now, let's prove that the sum of four consecutive whole numbers is divisible by 2.

Let's call the first number in the sequence "n". So, the four consecutive numbers will be n, n+1, n+2, and n+3.

To find the sum, we add these four numbers together:
n + (n+1) + (n+2) + (n+3)

If we combine all the "n" terms together, we get:
(4n + 6)

Now, let's use the property of divisibility by 2. An even number can be expressed as 2 multiplied by another whole number. Let's call this whole number "k".

So, an even number can be written as:
2k

Now, we need to prove that the sum (4n + 6) can be expressed as an even number (2k).

Let's substitute 4n + 6 with 2k:
4n + 6 = 2k

If we divide both sides of the equation by 2, we get:
2n + 3 = k

Now, we can clearly see that k is also a whole number since it's an integer multiplied by 2.

Hence, we have proved that the sum of four consecutive whole numbers, represented by (4n + 6), is divisible by 2.

To show that the sum of four consecutive whole numbers is divisible by 2, we can use the property of even and odd numbers.

Let's consider four consecutive whole numbers, which we can represent as n, n+1, n+2, and n+3.

To prove that their sum is divisible by 2, we need to show that the sum is an even number.

We know that an even number can be expressed as 2k, where k is an integer. Similarly, an odd number can be expressed as 2k+1.

Now, let's compute the sum of the four consecutive whole numbers:
n + (n+1) + (n+2) + (n+3)

Combining like terms, the sum becomes:
4n + 6

Using the fact that an even number can be expressed as 2k, we can rewrite the sum as:
2(2n + 3)

Here, 2n + 3 is an integer because n is an integer.

Thus, we have expressed the sum of the four consecutive whole numbers as twice an integer, which means the sum is divisible by 2.

Therefore, the sum of four consecutive whole numbers is indeed divisible by 2.

3 + 4 + 5 + 6 = 18

18/2 = 9

let the 4 numbers be

x, x+1, x+2, x+3 , where x is an integer.

x + x+1 + x+2 + x+3
= 4x + 6
= 2(x+3)

Any integer multiplied by 2 must be even
or
If any integer has a factor of 2, it must be even by definition.

The gcf(a,b) = 495 and lcm( a,b) =31,185 Find possible values of a and b if a is divisible by 35 and b is divisible by 81.