1-fizzbuzz.exs

## 1-fizzbuzz.exs
# Shortest (91 chars):
for n<-1..?d,r=&(rem(n,&1)==0&&IO.write&2)do;f=r.(3,"Fizz");b=r.(5,"Buzz");f||b||r.(1,n)end

# If newlines are required (99 chars):
for n<-1..?d,r=&(rem(n,&1)==0&&IO.write&2),do: (f=r.(3,"Fizz");b=r.(5,"Buzz");f||b||r.(1,n);:io.nl)

# Other solutions:
for n<-1..?d,r=&(rem(n,&1)==0&&IO.puts&2),do: r.(15,"FizzBuzz")||r.(3,"Fizz")||r.(5,"Buzz")||r.(1,n)
for n<-1..?d,do: Enum.find [{15,"FizzBuzz"},{3,"Fizz"},{5,"Buzz"},{1,n}],fn({d,t})->rem(n,d)==0&&IO.puts(t)end

## 2-notes.md

      
        

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?d is 100 (ASCII code for d)
for …,this,do abuses a for-comprehension filter. this runs on each iteration
and would skip if returning a falsy value.
Since IO.write returns a truthy value, we can chain f||b||r.(1,n) to only output
the bare number if Fizz or Buzz did not output.