Five times the smallest of three consecutive odd integers is ten more than twice the largest. Find the integers.
EDIT: Mind you, I just need help with setting it up. I can solve it, I'm just having a tough time writing it down properly.
Hello, Mad!
Five times the smallest of three consecutive odd integers
is ten more than twice the largest. Find the integers.
Consecutive odd integers "go up by 2's".
Let . $\displaystyle \begin{array}{ccc}x &=& \text{1st number} \\ x+2 &=& \text{2nd number} \\ x+4 &=& \text{3rd number}\end{array}$
$\displaystyle \underbrace{\text{Five times smallest}}\;\;\underbrace{\text{is}}\;\;\underbrac e{\text{twice largest}}\;\underbrace{\text{plus 10}}$
. . . . $\displaystyle 5 \times x \qquad\qquad =\qquad2(x+4)\qquad + 10$
And there is your equation: .$\displaystyle 5x \;=\;2(x+4) + 10$
You are correct. (Good spot! )
So let's redo this with the smallest number being 2x + 1 and the largest being 2x + 5:
$\displaystyle 5(2x + 1) = 2(2x + 5) + 10$
$\displaystyle 10x + 5 = 4x + 10 + 10$
$\displaystyle 6x = 15$
As there is no integer solution for x the problem does not have a solution. (In fact you can figure this out just by doing an "even-odd" argument: 5 times an odd number is odd; 2 times an odd number plus 10 is even. So the two sides can't be equal.)
-Dan