how many distinct values does it have? I solved this problem by manually adding parentheses and counting them, and got correct answer of 32. Is there a simple to find the answer? Thanks.

The value of the expression $1÷2÷3÷5÷7÷11÷13$ can be altered by including parentheses. If we are allowed to place as many parentheses as we want, how many distinct values can be obtained for this expression?

 A: In general, inserting parentheses in
 $$ a_1 \div a_2 \div a_3 \div \cdots \div a_n $$
can produce every number of the form
$$ a_1^{\strut}a_2^{-1}a_3^{s_3} a_4^{s_4} \cdots a_n^{s_n}$$
(and only those), where each $s_i$ is either $1$ or $-1$. Note that the exponents of $a_1$ and $a_2$ are fixed.
If $a_3$ through $a_n$ are coprime (as is the case here), this gives $2^{n-2}$ different possible values.
Proof by induction on $n$. The base case is $n=2$ where there is only a single possibility.
For $n>2$, first parenthesize $a_1\div\cdots\div a_{n-1}$ in order to get the desired exponents for $a_1$ through $a_{n-1}$.
Then, if the desired $s_n$ is opposite to $s_{n-1}$ then just replace $a_{n-1}$ by $(a_{n-1}\div a_n)$.
Otherwise the desired exponents $s_n$ and $s_{n-1}$ are equal. Let $E$ be the fully parenthesized expression such that $(E\div a_{n-1})$ is a subexpression of what we got from the induction hypothesis (that is, $E$ is the left operand to the division whose immediate right operand is $a_{n-1}$, and such a division always exists because $n-1\ge 2$), and replace this entire subexpression with $((E \div a_{n-1})\div a_n)$.
A: adding parenthesis changes the answer at all points but the end point due to the orders of operation. we have 6 divisions and one doesn't count so 5. each parentheses can either be included or not included so 2 options. in the end we have $2^5=32$
