Euler_Problem-030.b93

v                   >55+/ v v    $$_v#!:-1               <<
>01>::9::*:***0\>:#<|#\+1\<>9::*:***>:::55+%::*:**\55+/v>|:
                    >$ -#v_^>$>+\:#<_vv/+55\**:*::%+55:<->^
   ^                   +1<           $>:55+%::*:**\55+/v+
                                     .v/+55\**:*::%+55:<+
                                     @>:55+%::*:**\55+/v+
                                      v/+55\**:*::%+55:<+
                                      >55+%::*:**\55+$++^


v


// Get DigitCount

          >55+/ v
"x" 0\>:#<|#\+1\<
          >$@


//   stack ^5

::*:**

// Get Upper Bound (-> stack)

                  >55+/ v
1>::9::*:***0\>:#<|#\+1\<>9::*:***
                  >$ -#v_^
 ^                   +1<


// Test numbers (stack -> stack) (0 == isP5 )

     ::55+%::*:**\55+/v>
     v/+55\**:*::%+55:<-
     >:55+%::*:**\55+/v+
     v/+55\**:*::%+55:<+
     >:55+%::*:**\55+/v+
     v/+55\**:*::%+55:<+
     >55+%::*:**\55+/++^


---------------------------------------

**Yes**. This is in fact the first program without *put* or *get* instructions. It operates completely on the stack. And that makes it **really** fast.

But *- to be fair -* the algorithm is pretty simple:

First get the upper bound for our later search, we search for a number where `digitcount(9^5 * n) <= n`

After our algorithm calculated that number (I resisted the urge to hard code `354294`) we test every number from 0 to limit and sum the fitting ones (there are only 6).  
`4150`, `4151`, `54748`, `92727`, `93084`, `194979`