Euler_Problem-005.b93

45*00p110p120p >10g20g*10p00g20g1+:20p`#v_10g.@ 
               ^_v#`g00p03+1:g03   <p031<         v <
                 >10g30g%          |1              <^        p01/g03g01< 
                                   >10g30g/40p150p  > 20g50g1+:50p` #v_^
                 ^                                < |!%g05g04        <
                                                   ^<

 [00] 20 (const)
 [10] value
 [20] current-mult (1 .. 20)
 [30] divisor (for reducing)
 [40] value/divisor
 [50] test-mult (1 .. current-mult)

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We start with a value of `1`.
Then we multiply one after another the numbers from `1` to `20` to our value (We call these `multiplicand` in the loop).

After each multiplications we search go through the numbers from `2` to `value` and search for a number `divisor` where

 - `value % divisor == 0`
 - `(value / divisor) % {0, multiplicand} == 0`

Then (after we found such a number) we can reduce the value with it (`value /= divisor`).

The reduction steps guarantees us that we find the **smallest** number and it prevents our *Int64* numbers from overflowing