1. ## Odd Integers

What is the product if the largest of three consecutive odd integers whose sum is 129 is multiplied by 4?

Is the right equation:

x + (x + 2) + 4(x + 4) = 129?

2. no, first you need to find three consecutive odd integers whose sum is 129.

let the three
consecutive odd integers be $(x-2),x,(x+2)$,
so

$(x-2)+x+(x+2)=129$

solving this, you'll get $x$, and the largest odd integers is $(x+2)$

3. ## but...

Originally Posted by dedust
no, first you need to find three consecutive odd integers whose sum is 129.

let the three consecutive odd integers be $(x-2),x,(x+2)$,
so

$(x-2)+x+(x+2)=129$

solving this, you'll get $x$, and the largest odd integers is $(x+2)$
I've always seen x, x + 2 and x + 4 used to represent consecutive odd integers. Where did (x - 2) come from?

4. if you have $x$ as the first number, then the second is $x+2$ but now if you have $x$ as the second number then $x-2$ is the first,..is it clear??

just the trick to solve the problem

5. You can use whatever expressions provided that they are odd consecutives relative to $x$. You could have :

$(x + 88)$ and $(x + 90)$ and $(x + 92)$

or

$(x - 192)$ and $(x - 190)$ and $(x - 188)$

6. Originally Posted by Bacterius
You can use whatever expressions provided that they are odd consecutives relative to $x$. You could have :

$(x + 88)$ and $(x + 90)$ and $(x + 92)$

or

$(x - 192)$ and $(x - 190)$ and $(x - 188)$
yes,..choose the number that make it easier to solve

7. ## Ok...

Originally Posted by dedust
if you have $x$ as the first number, then the second is $x+2$ but now if you have $x$ as the second number then $x-2$ is the first,..is it clear??

just the trick to solve the problem
I thank you for this great trick and short cut. When I tried using my original equation, I ended up with 111/6, which didn't make sense to me at all.

8. ## Ray...

Originally Posted by Bacterius
You can use whatever expressions provided that they are odd consecutives relative to $x$. You could have :

$(x + 88)$ and $(x + 90)$ and $(x + 92)$

or

$(x - 192)$ and $(x - 190)$ and $(x - 188)$
Can we also use x, x + 2 and x + 4 as the list of consecutive odd integers to solve this question? When I did the algebra, I ended up with x = 111/6, which did not make sense to me at all.

9. Sure you can, but you got to rearrange your solution according to this at the end of your operations

10. ## Ray...

Originally Posted by Bacterius
Sure you can, but you got to rearrange your solution according to this at the end of your operations
Can you demonstrate for me?

11. Think about the meaning of $x$, and you should find out where something goes wrong

12. Originally Posted by sologuitar
Can we also use x, x + 2 and x + 4 as the list of consecutive odd integers to solve this question? When I did the algebra, I ended up with x = 111/6, which did not make sense to me at all.

how do you get x = 111/6 ?

Spoiler:
$x + (x+2) + (x+4) = 129$
$3x + 6 = 129$
$3x=123$
$x=41$

13. ## I see...

Originally Posted by dedust
how do you get x = 111/6 ?

Spoiler:
$x + (x+2) + (x+4) = 129$
$3x + 6 = 129$
$3x=123$
$x=41$
I made simple algebra errors that led to the wrong answer for x. Thanks for showing me the right way to find x.