A set of 4 consecutive odd integers sums to -256. Which integers are they?

Question

Please help! I am struggling!

Answer 1

Consecutive odd integers differ by 2, so they can be represented as n, n+2, n+4, etc.

Suppose there were 3 consecutive odd integers whose sum was -57.

Then the equation would be n + (n+2) + (n+4) = -57 , or

3n + 6 = -57

3n = -63

n = -21

So the 3 consecutive odd integers would be -21, -19 and -17.

Give it a try with your problem.

Answer 2

Let the smallest of the consecutive odd integers be "n". Then the next one is "n+2" (consecutive ODD integers, remember). The next two would be "n+4" and "n+6".

Their sum n+(n+2)+(n+4)+(n+6), which we know from the problem is "-256".

So, n+(n+2)+(n+4)+(n+6) = -256

Rearrange the addends (using the associative and commutative properties) to group like terms:

n+n+n+n+2+4+6 = -256

Add them up:

4n+12 = -256

Now we want to know "n", so we isolate it. It's like finding the center of an onion: we remove the outer layers one by one until we get there. We need to get rid of the 4 and the 12. The 4 is closest (by order of operations), so we take away the 12 by subtracting it (because it is added in). Since it is an equation, we must do the same on both sides.

4n+12-12 = -256-12

4n = -268

Next, we get rid of the 4 by dividing (since it's multiplied).

4n/4 = -268/4

n = -67

Now, we said that "n" is the smallest of the consecutive odd integers, so the others are found by adding 2 repeatedly.

So, the integers are -67, -65, -63, and -61.

We can check by finding their sum directly: -67 + -61 = -128, -65 + -63 = -128, -128 + -128 = -256

Answer 3

Let x=1st integer

x+2 = 2nd integer

x+4 = 3rd integer

x+6 = 4th integer

The sum of these is -256:

x+x+2+x+4+x+6 = -256 Combine like terms

4x+12=-256 Subtract 12 from both sides of the equation

4x+12-12=-256-12 Simplify

4x=-268 Divide both sides by 4

4x/4=-268/4

x=-67

x+2=-65

x+4=-63

x+6=-61