Precalculus

[danimalswim]

Zollie found four consecutive odd integers such that 5 times the sum of the first and the third was 22 greater than the product of 8 and the sum of the second and the fourth. What were the two integers?

 

[JeffM]

Zollie found four consecutive odd integers such that 5 times the sum of the first and the third was 22 greater than the product of 8 and the sum of the second and the fourth. What were the two integers?

TWO integers? The problem seems to be asking you to find four integers. What am I missing

The general formula for an odd integer is

\(\displaystyle 2n - 1.\)

So the next consecutive one would be

\(\displaystyle (2n - 1) + 2 = 2n + 1.\)

And the next would be? And the one after that?

You good to go now? You are looking for an equation in n.

 

[HallsofIvy]

Zollie found four consecutive odd integers such that 5 times the sum of the first and the third was 22 greater than the product of 8 and the sum of the second and the fourth. What were the two integers?

Four consecutive odd integers must be of the form 2n+ 1, 2n+ 3, 2n+ 5, 2n+ 7. "5 times the sum of the first and the third" is 5((2n+1)+ (2n+5))= 5(4n+ 6)= 20n+ 30. "The product of 8 and the sum of the second and the fourth" is 8(2n+3+ 2n+7)= 8(4n+10)= 32n+ 80. "22 greater" than that is 32n+ 80+ 22= 32n+ 102 so we must have 20n+ 30= 32n+ 102. However, there is NO integer n that satisfies that equation. That fact, together with your "What were the two integers?" when there were four integers referred to, makes me wonder if you have copied the problem correctly.

 

[JeffM]

Halls meant to say that there is no positive integer that satisfies that equation. The fact remains that we need to be sure that you have given us the problem correctly. That "two integers" does not inspire confidence.

PS: The difference between my general formula for an odd integer and Hall's is meaningless: two ways to say the same thing.

 

[lookagain]

PS: The difference between my general formula for an odd integer and Hall's is meaningless:
two ways to say the same thing.

danimalswim,

where there is a doable exercise, then it is sufficient to label the consecutive odd integers ** as:


Let n = the 1st (smallest) odd integer

Let n + 2 = the 2nd odd integer

Let n + 4 = the 3rd odd integer

Let n + 6 = the 4th (largest) odd integer



A simpler exercise example of mine:

"The sum of four consecutive odd integers is equal to -8. What are these four consecutive odd integers?"



Here is a beginning of a solution:


n + (n + 2) + (n + 4) + (n + 6) = -8


Then continue solving for n. After that, you can determine the next three larger odd integers.




** This approach works as well for a list of consecutive even integers.