Strategy to solve 'Insert Brackets' questions

Question

Is there some sort of strategy to work out the 'Insert Brackets' questions.
For example, how would one solve this?

Insert two sets of brackets into the following statement to make it true?
$$7+8-3\times 4-1\div 2=13$$

Is there a sort of strategy that works with this sorts of questions. Thanks

P.S. Please don't include things like limits in the answers. I won't understand it. I'm fine with Algebra I though

Answer 1

I don't think there is any general approach, just use your logical skills. In this given example $(7+(8-3)\times 4-1)\div 2=13$

Answer 2

Unless you have a ")" after the "1", you'd subtract $\frac12$ from an expression that is an integer, no matter how you put parentheses into the first part. So your must have

• $(\color{red}{7+8-3\times 4-1})\div2=13$,
• $7+(\color{red}{8-3\times 4-1})\div 2=13$,
• $7+8-(\color{red}{3\times 4-1})\div 2=13$, or
• $\color{red}{7+8-3}\times(4-1)\div2=13$,

where another pair of parentheses might be sprinkled into the red part. The only ways to do so in a way that actually affects the order of evaluation are

• $({(7+8-3)\times 4-1})\div2$, $({7+(8-3)\times 4-1})\div2$, $({7+8-3\times (4-1)})\div2$,
• $7+({(8-3)\times 4-1})\div 2$, $7+({8-3\times (4-1)})\div 2$
• $7+8-({3\times (4-1)})\div 2$,
• ${(7+8-3)}\times(4-1)\div2$, ${7+(8-3)}\times(4-1)\div2$

In most cases, you ultimately divide an odd number by $2$ and thus cannot reach the integer $13$ in the end

Answer 3

The expression has $n$ operators (here $n=5$), and placing parenthesis amounts to specifying in what order you perform them. So there are $n!$ ways to evaluate the expression (in case of $k$ commutative operators next to each other, order does not matter and you can simplify by $k!$).

You can maybe find some shortcuts such as "the dividend must be a multiple of the divisor", but I don't think that this reduces the number of trial in an essential way, and the search is essentially exhaustive (brute force).