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I was simply translating my Perl5 one-liners on ProjectEuler into Perl6's code.

These are all on its first problem:
```If we list all the natural numbers below 10 that are multiples of 3 or
+ 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

(I generalized this to:
below 10**\$n for code #1 and #2,
or below \$N for code #3)
This is my fastest solution in Perl5: (not in the code #1..4 above)
perl5 -e '\$n=3;print 2,3x--\$n,1 .6x\$n+2'

The following is the origin of code #1, which works beyond the limit of 64 bit int, since it uses string op:
\$n=3;print 2,(3x--\$n.1 .6x--\$n.8)=~s/^18/3/r

This is the origin of code #2:
\$n=3;\$_=2 .3x--\$n.1 .6x\$n;substr(\$_,-1)+=2;print

So code #3 is a more general solution for \$N which is typically not an integer exponentiation of 10:
map{\$s+=int\$_*(\$i=abs int 999/\$_)*++\$i/2}(3,5,-15);print\$s

They all give the same result: 233168 for \$n=3, but work for other \$n as well.
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