Question

Translate the following situation into an equation. Do not solve.

“The sum of three consecutive odd integers is 85 less than four times the first. What is the first integer?

Answer 1

[x+(x+2)+(x+4)] - [4x] =85
where x is the first integer

Answer 2

oh, I see. I would have not got that. Thanks!

Answer 3

In a more rigurous way
\[4n-(n+(n+2)+(n+4))=85 \leftarrow \rightarrow 4n-(3n+6)=85,n \in \mathbb{N}, n/2 \notin \mathbb{N} \]
AND remember that the sum is LESS than four times, the first answer is wrong, beause
\[n+n+2+n+4<4n \leftarrow \rightarrow (n+n+2+n+4)-4n <0\]

Answer 4

so what is the answer

Answer 5

so [x+(x+2)+(x+4)] - [4x] =85 is not part of the answer?

Answer 6

No, because that would mean the sum of the consecutive odd numbers is greater than four times the first odd number.

Answer 7

My apologies.
[4x] - [x+(x+2)+(x+4)] =85 is the required equation

Answer 8

I'd advise not to use x, usually it's implicit that
\[x \in \mathbb{R}\]
Whereas the solution is in |N

Answer 9

Let n be the middle number of 3 consecutive odd numbers. Then:
\[(n-2)+ n+(n+2)\text{=}4(n-2)-85\]