I was simply translating my Perl5 oneliners 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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