Order of operations

mcallendar

New member
Where would I put the parenthesis in this problem.

6+3/3+2-1-3=3

Staff member

Dr.Peterson

Elite Member
Where would I put the parenthesis in this problem.

6+3/3+2-1-3=3
Presumably you are asking where you can put parentheses in that equation to make it true. (The equation is not a problem; what you want to do with it is.)

All you can do is to try something; there is no magic or formula for this sort of thing. It's just meant to give you practice in evaluating, and perhaps to discover some general principles.

What have you tried?

I will tell you that I haven't yet seen an answer, so it may either be impossible, or use some unexpected trick. (Or maybe I just missed something obvious.)

lookagain

Senior Member
Where would I put the parenthesis in this problem.

6+3/3+2-1-3=3

It's been more than a week. I'm posting my full solutions for people to consider.

$$\displaystyle 6(+ \ 3/3)(+ \ 2 - 1) - 3 \ = \ 3$$

$$\displaystyle 6+ (3/3)(+ \ 2 - 1)( - 3) \ = \ 3$$

$$\displaystyle 6(+ \ (3/3)(+ \ 2) - 1) - 3 \ = \ 3$$

$$\displaystyle (6(+ \ 3)/3)(+ \ 2 - 1) - 3 \ = \ 3$$

$$\displaystyle 6 + ((3/3)(+ \ 2) - 1)(-3) \ = \ 3$$

$$\displaystyle 6(+ \ 3/(3(+ \ 2 - 1))) - 3 \ = 3$$

$$\displaystyle 6 + (3/(3(+ \ 2 - 1)))(-3) \ = \ 3$$

Last edited:

lookagain

Senior Member
I would like to add this to my list:

$$\displaystyle 6(+ \ 3/3) + (2 - 1)(- 3) \ = \ 3$$