Instructions

Test suite

Solution

Convert a number to a string, the contents of which depend on the number's factors.

- If the number has 3 as a factor, output 'Pling'.
- If the number has 5 as a factor, output 'Plang'.
- If the number has 7 as a factor, output 'Plong'.
- If the number does not have 3, 5, or 7 as a factor, just pass the number's digits straight through.

- 28's factors are 1, 2, 4,
**7**, 14, 28.- In raindrop-speak, this would be a simple "Plong".

- 30's factors are 1, 2,
**3**,**5**, 6, 10, 15, 30.- In raindrop-speak, this would be a "PlingPlang".

- 34 has four factors: 1, 2, 17, and 34.
- In raindrop-speak, this would be "34".

To run the tests, run the command `busted`

from within the exercise directory.

For more detailed information about the Lua track, including how to get help if you're having trouble, please visit the exercism.io Lua language page.

A variation on a famous interview question intended to weed out potential candidates. http://jumpstartlab.com

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
local raindrops = require('raindrops')
describe('raindrops', function()
it('the sound for 1 is 1', function()
assert.equal('1', raindrops(1))
end)
it('the sound for 3 is Pling', function()
assert.equal('Pling', raindrops(3))
end)
it('the sound for 5 is Plang', function()
assert.equal('Plang', raindrops(5))
end)
it('the sound for 7 is Plong', function()
assert.equal('Plong', raindrops(7))
end)
it('the sound for 6 is Pling as it has a factor 3', function()
assert.equal('Pling', raindrops(6))
end)
it('2 to the power 3 does not make a raindrop sound as 3 is the exponent not the base', function()
assert.equal('8', raindrops(8))
end)
it('the sound for 9 is Pling as it has a factor 3', function()
assert.equal('Pling', raindrops(9))
end)
it('the sound for 10 is Plang as it has a factor 5', function()
assert.equal('Plang', raindrops(10))
end)
it('the sound for 14 is Plong as it has a factor 7', function()
assert.equal('Plong', raindrops(7))
end)
it('the sound for 15 is PlingPlang as it has a factor 3 and 5', function()
assert.equal('PlingPlang', raindrops(15))
end)
it('the sound for 21 is PlingPlong as it has factors 3 and 7', function()
assert.equal('PlingPlong', raindrops(21))
end)
it('the sound for 25 is Plang as it has a factor 5', function()
assert.equal('Plang', raindrops(25))
end)
it('the sound for 27 is Pling as it has a factor 3', function()
assert.equal('Pling', raindrops(27))
end)
it('the sound for 35 is PlangPlong as it has factors 5 and 7', function()
assert.equal('PlangPlong', raindrops(35))
end)
it('the sound for 49 is Plong as it has a factor 7', function()
assert.equal('Plong', raindrops(49))
end)
it('the sound for 52 is 52', function()
assert.equal('52', raindrops(52))
end)
it('the sound for 105 is PlingPlangPlong as it has factors 3, 5 and 7', function()
assert.equal('PlingPlangPlong', raindrops(105))
end)
it('the sound for 3125 is Plang as it has a factor 5', function()
assert.equal('Plang', raindrops(3125))
end)
end)
```

```
return function(number)
local result
result = (number%3 == 0 and 'Pling' or '') .. (number%5 == 0 and 'Plang' or '') .. (number%7 == 0 and 'Plong' or '')
return #result == 0 and tostring(number) or result
end
```

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## Community comments

parens around the anded components may have helped me out with the intended logic bug I figured it out. Interesting approach!

@schecky I agree that the foo and bar or baz pattern looks a bit weird, but after a while you stop trying to parse out the logic and just recognize the pattern as a rough ternary operator.