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Daily Challenge Post #20 - Number Check

Create a function which checks a number for three different properties.

Is the number prime?
Is the number even?
Is the number a multiple of 10?

Each should return either true or false, which should be given as an array.

Today's challenge comes from sohilpandya on Codewars.


Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (13)

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gypsydave5 profile image
David Wickes • Edited

Well, as this isn't the most challenging challenge, I decided to challenge myself ;)

Here is an answer in the stack-based programming language Factor.

! is an integer even?
: even? ( int -- ? )
    2 mod 0 = ;

! is an integer a multiple of ten?
: multiple-of-ten? ( int -- ? )
    10 mod 0 = ;

! helper word (that's a 'function' in factor) for prime? 
: (prime?) ( int m -- ? )
    2dup <= [ t 2nip ] [ 
        2dup mod 0 = [ f 2nip ] [ 1 + (prime?) ] if
    ] if ;

! is an integer prime?
: prime? ( int -- ? )
    2 (prime?)

! all three tests applied to a single integer, output as an array
: all-three ( int -- seq )
    2dup
    [ prime? ] [ even? ] [ multiple-of-ten? ]
    tri* 3array ;


5 all-three
!
! --- Data stack:
! { t f f }
clear
10 all-three
!
! --- Data stack:
! { f t t }

I've never tried anything like this before, so I'd be interested in some feedback if any is available.

Conversely, if anyone would like me to explain what on earth is going on above, please ask and I'll do my best. I really enjoyed writing it.

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alvaromontoro profile image
Alvaro Montoro

JavaScript

const primeEvenTenth = number => {
  let isPrime = number > 1;
  for (let x = 2; x < Math.sqrt(number) && isPrime; x++)
    if (number % x === 0)
      isPrime = false;

  return [
    isPrime,
    number % 2 === 0,
    number % 10 === 0
  ]
}

Live demo on CodePen.

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jacobmgevans profile image
Jacob Evans

Well done!

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coreyja profile image
Corey Alexander • Edited

Rust Solution!

Went with the naive algorithm for testing for a prime number. Could always improve on that if needed :shruggy:

My test cases for these are also getting thinner and thinner as we go on...

fn is_prime(num: u64) -> bool {
    if num == 0 {
        return false;
    }

    for i in 2..num {
        if num % i == 0 {
            return false;
        }
    }

    true
}

pub fn number_property(i: i64) -> (bool, bool, bool) {
    let is_even = i % 2 == 0;
    let is_divisible_by_ten = i % 10 == 0;

    let is_prime = if i > 0 { is_prime(i as u64) } else { false };

    (is_prime, is_even, is_divisible_by_ten)
}

#[cfg(test)]
mod tests {
    use crate::number_property;

    #[test]
    fn it_works_for_the_examples() {
        assert_eq!(number_property(7), (true, false, false));
        assert_eq!(number_property(10), (false, true, true));
    }

    #[test]
    fn it_works_for_negative_examples() {
        assert_eq!(number_property(-7), (false, false, false));
        assert_eq!(number_property(-10), (false, true, true));
    }
}
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jacobmgevans profile image
Jacob Evans

Nice!

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choroba profile image
E. Choroba

Perl solution, using the modulo operator.

#!/usr/bin/perl
use warnings;
use strict;

sub prime_even_10 {
    my ($x) = @_;
    return $x > 1 && !(grep 0 == $x % $_, 2 .. sqrt $x),
           0 == $x % 2,
           0 == $x % 10
}

use Test::More tests => 11;

is_deeply [prime_even_10(0)],  [!1,  1,  1];
is_deeply [prime_even_10(1)],  [!1, !1, !1];
is_deeply [prime_even_10(2)],  [ 1,  1, !1];
is_deeply [prime_even_10(3)],  [ 1, !1, !1];
is_deeply [prime_even_10(4)],  [!1,  1, !1];
is_deeply [prime_even_10(5)],  [ 1, !1, !1];
is_deeply [prime_even_10(6)],  [!1,  1, !1];
is_deeply [prime_even_10(7)],  [ 1, !1, !1];
is_deeply [prime_even_10(8)],  [!1,  1, !1];
is_deeply [prime_even_10(9)],  [!1, !1, !1];
is_deeply [prime_even_10(10)], [!1,  1,  1];
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mannyluvstacos profile image
Manny

:) I really enjoyed this.
It was a nice way to start the morning.
I did not know this existed and now I will look forward to this each morning to start my day or to give a break during the day <3
image

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peter279k profile image
peter279k

Here is my simple solution with Python:

import math

def is_prime(n):
    if n == 2:
        return True
    if n == 1 or n <= 0:
        return False
    number = 2
    end_number = math.sqrt(n)

    while number <= end_number:
        if n % number == 0:
            return False
        number += 1

    return True

def is_even(n):
    return n % 2 == 0

def is_multiple_ten(n):
    return n % 10 == 0

def number_property(n):
    answer = [is_prime(n), is_even(n), is_multiple_ten(n)]

    return answer
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mrdulin profile image
official_dulin

Go:

package kata

func NumberProperty(n int) []bool {
  return []bool{isPrime(n), isEven(n), multipleOfTen(n)}
}

func isPrime(n int) bool {
  for i := 2; i < n; i ++ {
    if n % i == 0 {
      return false
    }
  }
  return n > 1
}

func isEven(n int) bool {
  return n % 2 == 0
}

func multipleOfTen(n int) bool {
  return n % 10 == 0
}
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brightone profile image
Oleksii Filonenko

Elixir:

defmodule NumberCheck do
  require Integer

  @spec check(number) :: {prime :: boolean, even :: boolean, multiple_of_ten :: boolean}
  def check(number), do: {prime?(number), Integer.is_even(number), rem(number, 10) == 0}

  @spec prime?(number) :: boolean
  defp prime?(n) when n < 2, do: false
  defp prime?(n), do: not Enum.any?(2..(n |> :math.sqrt() |> floor()), &(rem(n, &1) == 0))
end
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kvharish profile image
K.V.Harish • Edited

My solution in js

const numberCheck = (num) => {
  return [`isPrime: ${(() => {
      const sqrtOfNum = Math.sqrt(num);
      for(let index = 2; index <= sqrtOfNum; index++) {
        if(num % index === 0) {
          return false;
        }
      } 
      return num > 1;
    })()}`,
    `isEven: ${num % 2 === 0}`,
    `isMultipleOfTen: ${num % 10 === 0}`];
};
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jacobmgevans profile image
Jacob Evans • Edited

I made the answer earlier and forgot to put it here after I solved it in CodeWars lmao
JavaScript answer, Big O time complexity of O(sqrt(n)) to find if N is a Prime Number.

const isPrime = num => {
    for(let i = 2, s = Math.sqrt(num); i <= s; i++)
        if(num % i === 0) return false; 
    return num > 1;
}

const numberProperty = n =>  [
   isPrime(n),
   n % 2 === 0,  
   n % 10 === 0,   
];
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kaspermeyer profile image
Kasper Meyer

Ruby 2.6

require "minitest/autorun"
require "prime"

class ArbitraryPropertyNumber
  def initialize number
    @number = number
  end

  def prime?
    Prime.prime?(@number)
  end

  def even?
    @number.even?
  end

  def multiple_of_ten?
    @number % 10 == 0
  end
end

class ArbitraryPropertyNumberValidator
  def initialize number
    @number = number
  end

  def validate
    [number.prime?, number.even?, number.multiple_of_ten?]
  end

  private

    def number
      ArbitraryPropertyNumber.new(@number)
    end
end

class ArbitraryPropertyNumberValidatorTest < MiniTest::Test
  def test_first_element_is_true_when_number_is_prime
    assert_equal [true, false, false], ArbitraryPropertyNumberValidator.new(3).validate
  end

  def test_second_element_is_true_when_number_is_even
    assert_equal [false, true, false], ArbitraryPropertyNumberValidator.new(4).validate
  end

  def test_third_element_is_true_when_number_is_multiple_of_ten
    assert_equal [false, true, true], ArbitraryPropertyNumberValidator.new(20).validate
  end
end