Math Help - 3 Consecutive Odd Integers

1. 3 Consecutive Odd Integers

What is the largest of three consecutive odd integers if the product of the first and third integers is 6 more than three times the second integer?

2. Originally Posted by magentarita
What is the largest of three consecutive odd integers if the product of the first and third integers is 6 more than three times the second integer?
Let the largest of the odd integers be $2x+1$, then the others are $2(x-1)+1$ and $2(x-2)+1$ (where $x$ is an integer).

So we are told that: $[2(x-2)+1][2x+1]=3[2(x-1)+1]+6$

This is a quadratic in $x$, so solve it and choose the root (or roots) that makes sense.

CB

3. Hello, magentarita!

Another approach . . .

What is the largest of three consecutive odd integers if the product
of the first and third integers is 6 more than three times the second integer?
Consecutive odd integers "go up by two's".

Let $x$ = the smallest odd integer.
Then $(x+2)\text{ and }(x+4)$ are the next two odd integers.

. . $\underbrace{\text{Product of first and third}} _{x(x+4)}\; \underbrace{\text{is}}_{=} \; \underbrace{\text{6 more than}}_{6\;\; +} \; \underbrace{\text{3 times second}}_{3(x+2)}$

And we have: . $x(x+4) \;=\;3(x+2)+6 \quad\Rightarrow\quad x^2 + x - 12 \:=\:0$

Solve the quadratic equation (and take the odd root).

And remember, they asked for the largest of the three.

4. ok.......

Originally Posted by CaptainBlack
Let the largest of the odd integers be $2x+1$, then the others are $2(x-1)+1$ and $2(x-2)+1$ (where $x$ is an integer).

So we are told that: $[2(x-2)+1][2x+1]=3[2(x-1)+1]+6$

This is a quadratic in $x$, so solve it and choose the root (or roots) that makes sense.

CB
Great work!

5. ok.........

Originally Posted by Soroban
Hello, magentarita!

Another approach . . .

Consecutive odd integers "go up by two's".

Let $x$ = the smallest odd integer.
Then $(x+2)\text{ and }(x+4)$ are the next two odd integers.

. . $\underbrace{\text{Product of first and third}} _{x(x+4)}\; \underbrace{\text{is}}_{=} \; \underbrace{\text{6 more than}}_{6\;\; +} \; \underbrace{\text{3 times second}}_{3(x+2)}$

And we have: . $x(x+4) \;=\;3(x+2)+6 \quad\Rightarrow\quad x^2 + x - 12 \:=\:0$

Solve the quadratic equation (and take the odd root).

And remember, they asked for the largest of the three.
I save your replies for future reference.