Dumb Algebra I Question

[Reefgazer]

OK, I am sure this question is simple, but I am unable to figure it out and offer DD an explanation. If a question asks us to find four consecutive odd integers, we have to indicate the integers as N, N+2, N+4, and N+6. Question is this: Doesn't wether N +2, N+4 or N+6 give us an odd integer depend on what you start with (N)? It's clear to me that you have to add 2 more than N to get to an odd integer, but whether or not you get to an odd or an even integer depends on what you start with (N), correct? I know I'm wording this poorly, but I am hoping someone understands what I'm asking and can explain why you always add 2 to N to get an even integer, and you always add 2 to N to get an odd integer, regardless of whether you know in advance what number you're starting with ( what number N is). OMG, that sounds so confusing, LOL.

[justasque]

OK, I am sure this question is simple, but I am unable to figure it out and offer DD an explanation. If a question asks us to find four consecutive odd integers, we have to indicate the integers as N, N+2, N+4, and N+6. Question is this: Doesn't wether N +2, N+4 or N+6 give us an odd integer depend on what you start with (N)? It's clear to me that you have to add 2 more than N to get to an odd integer, but whether or not you get to an odd or an even integer depends on what you start with (N), correct? I know I'm wording this poorly, but I am hoping someone understands what I'm asking and can explain why you always add 2 to N to get an even integer, and you always add 2 to N to get an odd integer, regardless of whether you know in advance what number you're starting with ( what number N is). OMG, that sounds so confusing, LOL.

Using N, N+2, etc will not guarantee you get an odd value for N. Yes, whether N+2 etc. gets you an odd integer does depend on whether N itself is odd. I am assuming that there is something else to the problem that will constrain it to only odd numbers, or that you will get multiple answers for N and will have to choose the odd one among them.

[maize]

Can you post the entire question?

[maize]

I could say: "let N be any odd number"

[Arcadia]

The pattern for the integer number line is

 

O(odd)E(even)OEOEOEOEOEOEOEOE

 

So if N is odd, N+2 gives you the next odd number.

And if N is even, N+2 gives you the next even number.

[Reefgazer]

The question in its entirety reads: Find four consecutive odd integers such that six times the sum of the first and the third is 3 more than 5 times the opposite of the fourth.

 

Using N, N+2, etc will not guarantee you get an odd value for N. Yes, whether N+2 etc. gets you an odd integer does depend on whether N itself is odd. I am assuming that there is something else to the problem that will constrain it to only odd numbers, or that you will get multiple answers for N and will have to choose the odd one among them.

[Reefgazer]

Yes this is exactly the problem I'm having explaining to DD. Neither one of us can understand how adding two to another number automatically gives you an odd integer unless you know in advance what N is. I posted the problem is and it's entirety below.

Using N, N+2, etc will not guarantee you get an odd value for N. Yes, whether N+2 etc. gets you an odd integer does depend on whether N itself is odd. I am assuming that there is something else to the problem that will constrain it to only odd numbers, or that you will get multiple answers for N and will have to choose the odd one among them.

[Reefgazer]

Oh, I'm in idiot. I think I just figured out this question. It is clear that I'm starting with an odd integer because it says four consecutive odd integers in the problem. Am I thinking that through right?

The question in its entirety reads: Find four consecutive odd integers such that six times the sum of the first and the third is 3 more than 5 times the opposite of the fourth.

 

 

[justasque]

The question in its entirety reads: Find four consecutive odd integers such that six times the sum of the first and the third is 3 more than 5 times the opposite of the fourth.

So, 6(N + (N + 4)) = 3+ 5*(-(N+6))

6(2N+4) = 3+ -5N -30

12N + 24 = -5N -27

17N = -51

N = -3

So

N+2 = -1

N+4 = 1

N+6 = 3

 

Check:

Find four consecutive odd integers such that six times the sum of the first and the third is 3 more than 5 times the opposite of the fourth.

6*(-3 + 1) = 3 + 5*-(3)

6*(-2) = 3 + (-15)

-12 = 3 + -15

-12 = -12

Check!

 

If you work through it, N does come out to be an odd number, although it was not specifically/overtly constrained to be odd. Had it come out to be even, the answer would be "no solultion". Had there been several values for N, you would have needed to select the odd one(s) to be the solution set.

[maize]

If you make this into an equation you will get only 1 answer.

[maize]

Ah, justesque wrote it all out above.

[justasque]

Yes this is exactly the problem I'm having explaining to DD. Neither one of us can understand how adding two to another number automatically gives you an odd integer unless you know in advance what N is. I posted the problem is and it's entirety below.

That's because adding two to another number does NOT automatically give you an odd integer (unless there is also some other way to constrain N to be odd.) It just ensures that IF N is odd, then the other three numbers are odd, and IF N is even, then the other three numbers are even.

[justasque]

Oh, I'm in idiot. I think I just figured out this question. It is clear that I'm starting with an odd integer because it says four consecutive odd integers in the problem. Am I thinking that through right?

No, just saying it in the problem doesn't make N come out to be odd. They seem to use "odd" to get you to use the N, N+2, etc. set-up. N does come out to be odd, but it's because the answer (to the equation) happens to be odd, not because you did something in the set-up of the problem to guarantee N is odd.

Edited March 4, 2016 by justasque

[Cosmos]

Sometimes when you solve a problem, you don't initially use all of the information given in the problem to find possible solutions. But then  you use that information at the end to either confirm that your solutions are valid or to eliminate invalid solutions. In this problem you used these pieces of information:

 

1. the four numbers are 2 units apart on the number line

2. six times the sum of the first and the third is 3 more than 5 times the opposite of the fourth

 

You did not use

 

3. the four numbers are odd

 

But after you got a possible solution, you needed to go back and check that it satisfied 3. As justaque explained, if they had come out even, you would then know that the problem does not have a valid solution.

[debi21]

I might not know what I'm saying as I'm sick, but... in this problem, as others have said, you don't need to constrain n to be odd - as long as n is odd, n+2, n+4 etc will be odd. If you did want or need to constrain it to be odd you could call the first number 2n+1, and subsequent numbers 2n+3, 2n+5. This guarantees they are odd, but complicates the arithmetic and your first number isn't the n you would solve for but still 2n+1.