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Published at Jun 07 2020
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Instructions

Test suite

Solution

Compute the prime factors of a given natural number.

A prime number is only evenly divisible by itself and 1.

Note that 1 is not a prime number.

What are the prime factors of 60?

- Our first divisor is 2. 2 goes into 60, leaving 30.
- 2 goes into 30, leaving 15.
- 2 doesn't go cleanly into 15. So let's move on to our next divisor, 3.

- 3 goes cleanly into 15, leaving 5.
- 3 does not go cleanly into 5. The next possible factor is 4.
- 4 does not go cleanly into 5. The next possible factor is 5.

- 5 does go cleanly into 5.
- We're left only with 1, so now, we're done.

Our successful divisors in that computation represent the list of prime factors of 60: 2, 2, 3, and 5.

You can check this yourself:

- 2 * 2 * 3 * 5
- = 4 * 15
- = 60
- Success!

For installation and learning resources, refer to the Ruby resources page.

For running the tests provided, you will need the Minitest gem. Open a terminal window and run the following command to install minitest:

```
gem install minitest
```

If you would like color output, you can `require 'minitest/pride'`

in
the test file, or note the alternative instruction, below, for running
the test file.

Run the tests from the exercise directory using the following command:

```
ruby prime_factors_test.rb
```

To include color from the command line:

```
ruby -r minitest/pride prime_factors_test.rb
```

The Prime Factors Kata by Uncle Bob http://butunclebob.com/ArticleS.UncleBob.ThePrimeFactorsKata

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

```
require 'minitest/autorun'
require_relative 'prime_factors'
class PrimeFactorsTest < Minitest::Test
def test_1
assert_equal [], PrimeFactors.of(1)
end
def test_2
skip
assert_equal [2], PrimeFactors.of(2)
end
def test_3
skip
assert_equal [3], PrimeFactors.of(3)
end
def test_4
skip
assert_equal [2, 2], PrimeFactors.of(4)
end
def test_6
skip
assert_equal [2, 3], PrimeFactors.of(6)
end
def test_8
skip
assert_equal [2, 2, 2], PrimeFactors.of(8)
end
def test_9
skip
assert_equal [3, 3], PrimeFactors.of(9)
end
def test_27
skip
assert_equal [3, 3, 3], PrimeFactors.of(27)
end
def test_625
skip
assert_equal [5, 5, 5, 5], PrimeFactors.of(625)
end
def test_901255
skip
assert_equal [5, 17, 23, 461], PrimeFactors.of(901_255)
end
def test_93819012551
skip
assert_equal [11, 9539, 894_119], PrimeFactors.of(93_819_012_551)
end
end
```

```
require 'Prime'
class PrimeFactors
def self.of(n)
raise DomainError if n <= 0
return [] if n == 1
# problem has no restriction on using library functions;
# get the prime factors & exponents; format it as expected
n.prime_division
.each_with_object([]) { |(x, exp), obj| obj << [x] * exp }
.flatten
end
end
```

Finally - a math problem that doesn't impose asinine restrictions on using standard library functionality.

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