Show that the sum of 3 consecutive odd integers is a multiple of 3?
I'm not so sure on how to answer this question, can you please help and show me on how to find it?
any three consecutive numbers:
x , x+1, x+2
their sum
= x + x+1 + x+2
= 3x + 3
= 3(x+1)
since it contains a factor of 3, it must be a multiple of 3
Look at Damon's correct solution to the one above this.
I did not read the question carefully enough and proved that the sum of any 3 consecutive numbers is a multiple of 3.
btw, that is also true for any 3 consecutive even numbers.
it's even easier to see if you call them
x-2, x, x+2
what's the sum?
Sure! To show that the sum of 3 consecutive odd integers is a multiple of 3, let's break down the problem step by step.
First, let's express the three consecutive odd integers in terms of a variable. We can choose any variable, but let's use "n" for simplicity.
So, let the first odd integer be "n," the second odd integer be "n + 2," and the third odd integer be "n + 4."
Now, let's find the sum of these three odd integers:
n + (n + 2) + (n + 4)
Combining like terms, we get:
3n + 6
To prove that this sum is a multiple of 3, we can use algebra by factoring. Since 3n is a multiple of 3, we can factor it out:
3(n + 2)
Now, we have the sum expressed as 3 times (n + 2). This shows that the sum is indeed a multiple of 3.
Therefore, the sum of 3 consecutive odd integers is always a multiple of 3.
I hope this breakdown helps you understand how to approach and solve this type of problem!