with prime factorization 
,
numbers that multiply to 
Y by
Hello, I was wondering if it's possible to make 8 with the numbers 5, 3, 5, and 7 under the following rules:
-You can only use 5 twice, 3 once, and 7 once.
-You must use all the numbers.
You can stack numbers to form larger numbers (example: I could take 3 and 5 and turn it into 35 or 53, or use 7, 3, and 5 to make 375.)
-You are allowed to use parentheses.
(Also, I already found out that no 3 digital numbers will work for the solution.)
-You can only use 5 twice, 3 once, and 7 once.
-You must use all the numbers.
You can stack numbers to form larger numbers (example: I could take 3 and 5 and turn it into 35 or 53, or use 7, 3, and 5 to make 375.)
-You are allowed to use parentheses.
(Also, I already found out that no 3 digital numbers will work for the solution.)
![\[|P\left(\left\lfloor \sqrt{\sqrt{\left\lceil \sqrt{\left\lfloor \sqrt{\left\lceil \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\left\lceil \sqrt{\sqrt{\sqrt{5537}}}\right\rceil !!}}\right\rfloor !}}\right\rceil !}\right\rfloor !}\right\rceil !}}\right\rfloor\right)|\]](http://latex.artofproblemsolving.com/8/7/b/87ba34a74433ea04a72832926fdbc4a618408d4c.png)
![\[
(5 + 5) - (7 - 3) = 8
\]](http://latex.artofproblemsolving.com/e/c/5/ec56494a3dde4ad303e38f5d9e94cb84428e1a51.png)
.
.
.
,
where | | denotes cardinality
,
is equal to 8, so he solved the problem, he is not using eight
