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def it(n) | |
x = [*2..n] | |
x.each do |i| | |
[*(i+1)..n].each { |j| x.delete(j) if (j%i).zero? } | |
end | |
end | |
# 1) what does it do? | |
# 2) how fast is it? | |
# 3) can you make it faster? | |
# | |
# Email answers to [email protected] with subject "Localytics Coding Challenge" |
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