diff --git a/src/Julia/Problems001-050/Problem001.jl b/src/Julia/Problems001-050/Problem001.jl
new file mode 100644
index 0000000..19de345
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem001.jl
@@ -0,0 +1,27 @@
+#=
+Created on 08 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 1 of Project Euler
+https://projecteuler.net/problem=1
+=#
+
+function Problem1()
+    #=
+    If we list all the natural numbers below 10 that are multiples of 3 or 5,
+    we get 3, 5, 6 and 9. The sum of these multiples is 23.
+
+    Find the sum of all the multiples of 3 or 5 below 1000.
+    =#
+    ans = sum(x for x in 0:999 if x%3==0 || x%5==0)
+    
+    return ans
+end
+
+
+println("Time to evaluate Problem 1:")
+@time Problem1()
+println("")
+println("Result for Problem 1: ", Problem1())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem002.jl b/src/Julia/Problems001-050/Problem002.jl
new file mode 100644
index 0000000..34bb301
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem002.jl
@@ -0,0 +1,40 @@
+#=
+Created on 08 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 2 of Project Euler
+https://projecteuler.net/problem=2
+=#
+
+function Problem2()
+    #=
+    Each new term in the Fibonacci sequence is generated by adding the
+    previous two terms. By starting with 1 and 2, the first 10 terms will be:
+
+    1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
+
+    Find the sum of all the even-valued terms in the sequence which do not
+    exceed four million.
+    =#
+
+    ans = 0
+    limit = 4_000_000
+    x, y = 1, 1
+    z = x + y  # Because every third Fibonacci number is even
+    while z <= limit
+        ans += z
+        x = y + z
+        y = z + x
+        z = x + y
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 2:")
+@time Problem2()
+println("")
+println("Result for Problem 2: ", Problem2())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem003.jl b/src/Julia/Problems001-050/Problem003.jl
new file mode 100644
index 0000000..af1da0a
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem003.jl
@@ -0,0 +1,34 @@
+#=
+Created on 15 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 3 of Project Euler
+https://projecteuler.net/problem=3
+=#
+
+function Problem3()
+    #=
+    The prime factors of 13195 are 5, 7, 13 and 29.
+
+    What is the largest prime factor of the number 600851475143 ?
+    =#
+
+    ans = 600_851_475_143
+    factor = 2
+    while factor * factor < ans
+        while ans % factor == 0
+            ans = ans ÷ factor
+        end
+        factor += 1
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 3:")
+@time Problem3()
+println("")
+println("Result for Problem 3: ", Problem3())
diff --git a/src/Julia/Problems001-050/Problem004.jl b/src/Julia/Problems001-050/Problem004.jl
new file mode 100644
index 0000000..aeba860
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem004.jl
@@ -0,0 +1,36 @@
+#=
+Created on 17 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 4 of Project Euler
+https://projecteuler.net/problem=4
+=#
+
+function Problem4()
+    #=
+    A palindromic number reads the same both ways. The largest palindrome made
+    from the product of two 2-digit numbers is 9009 = 91 x 99.
+
+    Find the largest palindrome made from the product of two 3-digit numbers.
+    =#
+    ans = 0
+    for i in 100:1000
+        for j in 100:1000
+            palindrome = i * j
+            s = string(palindrome)
+            if (s==reverse(s)) & (palindrome > ans)
+                ans = palindrome
+            end
+        end
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 4:")
+@time Problem4()
+println("")
+println("Result for Problem 4: ", Problem4())
diff --git a/src/Julia/Problems001-050/Problem005.jl b/src/Julia/Problems001-050/Problem005.jl
new file mode 100644
index 0000000..b61d5fb
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem005.jl
@@ -0,0 +1,41 @@
+#=
+Created on 20 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 5 of Project Euler
+https://projecteuler.net/problem=5
+=#
+
+#=
+The LCM of two natural numbers x and y is given by:
+def lcm(x, y):
+    return x * y // math.gcd(x, y)
+
+It is possible to compute the LCM of more than two numbers by iteratively
+computing the LCM of two numbers, i.e. LCM(a, b, c) = LCM(a, LCM(b, c))
+=#
+
+
+function Problem5()
+    #=
+    2520 is the smallest number that can be divided by each of the numbers
+    from 1 to 10 without any remainder.
+
+    What is the smallest positive number that is evenly divisible by all of
+    the numbers from 1 to 20?
+    =#
+    ans = 1
+    for i in 1:21
+        ans *= i ÷ gcd(i, ans)
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 5:")
+@time Problem5()
+println("")
+println("Result for Problem 5: ", Problem5())
diff --git a/src/Julia/Problems001-050/Problem006.jl b/src/Julia/Problems001-050/Problem006.jl
new file mode 100644
index 0000000..e923edd
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem006.jl
@@ -0,0 +1,36 @@
+#=
+Created on 20 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 6 of Project Euler
+https://projecteuler.net/problem=6
+=#
+
+function Problem6()
+    #=
+    The sum of the squares of the first ten natural numbers is,
+    1^2 + 2^2 + ... + 10^2 = 385
+
+    The square of the sum of the first ten natural numbers is,
+    (1 + 2 + ... + 10)^2 = 55^2 = 3025
+
+    Hence the difference between the sum of the squares of the first ten
+    natural numbers and the square of the sum is 3025 − 385 = 2640.
+
+    Find the difference between the sum of the squares of the first one
+    hundred natural numbers and the square of the sum. Statement
+    =#
+    n = 100
+    square_of_sum = sum(i for i in (1:n))^2
+    sum_squares = sum(i^2 for i in 1:n)
+    diff = square_of_sum - sum_squares
+    return diff
+end
+
+
+println("Time to evaluate Problem 6:")
+@time Problem6()
+println("")
+println("Result for Problem 6: ", Problem6())
diff --git a/src/Julia/Problems001-050/Problem007.jl b/src/Julia/Problems001-050/Problem007.jl
new file mode 100644
index 0000000..30f9539
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem007.jl
@@ -0,0 +1,36 @@
+#=
+Created on 24 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 7 of Project Euler
+https://projecteuler.net/problem=7
+=#
+
+using Primes
+
+function Problem7()
+    #=
+    By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13,
+    we can see that the 6th prime is 13.
+
+    What is the 10_001st prime number
+    =#
+    number = 2
+    primeList = Int[]
+    while length(primeList) < 10_001
+        if isprime(number)
+            append!(primeList,number)
+        end
+        number += 1
+    end
+
+    return primeList[length(primeList)]
+end
+
+
+println("Time to evaluate Problem 7:")
+@time Problem7()
+println("")
+println("Result for Problem 7: ", Problem7())
diff --git a/src/Julia/Problems001-050/Problem008.jl b/src/Julia/Problems001-050/Problem008.jl
new file mode 100644
index 0000000..a2c5397
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem008.jl
@@ -0,0 +1,68 @@
+#=
+Created on 29 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 8 of Project Euler
+https://projecteuler.net/problem=8
+=#
+
+function Problem8()
+    #=
+    The four adjacent digits in the 1000-digit number that have the
+    greatest product are 9 × 9 × 8 × 9 = 5832.
+
+            731671...963450
+
+    Find the thirteen adjacent digits in the 1000-digit number that have
+    the greatest product. What is the value of this product?
+    =#
+
+    NUM = """
+        73167176531330624919225119674426574742355349194934
+        96983520312774506326239578318016984801869478851843
+        85861560789112949495459501737958331952853208805511
+        12540698747158523863050715693290963295227443043557
+        66896648950445244523161731856403098711121722383113
+        62229893423380308135336276614282806444486645238749
+        30358907296290491560440772390713810515859307960866
+        70172427121883998797908792274921901699720888093776
+        65727333001053367881220235421809751254540594752243
+        52584907711670556013604839586446706324415722155397
+        53697817977846174064955149290862569321978468622482
+        83972241375657056057490261407972968652414535100474
+        82166370484403199890008895243450658541227588666881
+        16427171479924442928230863465674813919123162824586
+        17866458359124566529476545682848912883142607690042
+        24219022671055626321111109370544217506941658960408
+        07198403850962455444362981230987879927244284909188
+        84580156166097919133875499200524063689912560717606
+        05886116467109405077541002256983155200055935729725
+        71636269561882670428252483600823257530420752963450
+        """
+
+    num = replace(NUM, "\n" => "")
+    adjacent_digits = 13
+    target = length(num) - adjacent_digits
+    ans, i = 0, 1
+    while i < target
+        a, j = 1, i
+        while j < (i + adjacent_digits)
+            a = a * parse(Int, (num[j]))
+            j += 1
+        end
+        if a > ans
+            ans = a
+        end
+        i += 1
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 8:")
+@time Problem8()
+println("")
+println("Result for Problem 8: ", Problem8())
diff --git a/src/Julia/Problems001-050/Problem009.jl b/src/Julia/Problems001-050/Problem009.jl
new file mode 100644
index 0000000..fc1bc62
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem009.jl
@@ -0,0 +1,38 @@
+#=
+Created on 01 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 9 of Project Euler
+https://projecteuler.net/problem=9
+=#
+
+function Problem9()
+    #=
+    A Pythagorean triplet is a set of three natural numbers, a < b < c,
+    for which a^2 + b^2 = c^2
+
+    For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+
+    There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+    Find the product abc.
+    =#
+
+    upper_limit = 1000
+    for a in 1:upper_limit + 1
+        for b in a + 1:upper_limit + 1
+            c = upper_limit - a - b
+            if a * a + b * b == c * c
+                # It is now implied that b < c, because we have a > 0
+                return a * b * c
+            end
+        end
+    end
+end
+
+
+println("Time to evaluate Problem 9:")
+@time Problem9()
+println("")
+println("Result for Problem 9: ", Problem9())
diff --git a/src/Julia/Problems001-050/Problem010.jl b/src/Julia/Problems001-050/Problem010.jl
new file mode 100644
index 0000000..988743d
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem010.jl
@@ -0,0 +1,27 @@
+#=
+Created on 03 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 10 of Project Euler
+https://projecteuler.net/problem=10
+=#
+
+using Primes
+
+function Problem10()
+    #=
+    The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
+
+    Find the sum of all the primes below two million.
+    =#
+
+    return sum(primes(1_999_999))
+end
+
+
+println("Time to evaluate Problem 10:")
+@time Problem10()
+println("")
+println("Result for Problem 10: ", Problem10())
diff --git a/src/Julia/Problems001-050/Problem012.jl b/src/Julia/Problems001-050/Problem012.jl
new file mode 100644
index 0000000..17054df
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem012.jl
@@ -0,0 +1,57 @@
+#=
+Created on 21 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 12 of Project Euler
+https://projecteuler.net/problem=12
+=#
+
+
+
+function Problem12()
+    #=
+     The sequence of triangle numbers is generated by adding the natural
+    numbers. So the 7th triangle number would be:
+    1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
+
+    The first ten terms would be:
+    1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+
+    Let us list the factors of the first seven triangle numbers:
+
+     1: 1
+     3: 1,3
+     6: 1,2,3,6
+    10: 1,2,5,10
+    15: 1,3,5,15
+    21: 1,3,7,21
+    28: 1,2,4,7,14,28
+
+    We can see that 28 is the first triangle number to have over five divisors.
+
+    What is the value of the first triangle number to have over five hundred
+    divisors?
+    =#
+
+    function num_divisors(n)
+        r = isqrt(n)
+        2 * count(n % i == 0 for i in 1:r) - (r^2 == n)
+    end
+    
+
+    triangle = 0
+    for i in Iterators.countfrom(1)
+        triangle += i
+        if num_divisors(triangle) > 500
+            return string(triangle)
+        end
+    end
+end
+
+
+println("Time to evaluate Problem 12:")
+@time Problem12()
+println("")
+println("Result for Problem 12: ", Problem12())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem013.jl b/src/Julia/Problems001-050/Problem013.jl
new file mode 100644
index 0000000..b8a28e8
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem013.jl
@@ -0,0 +1,27 @@
+using Base: String
+#=
+Created on 22 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 13 of Project Euler
+https://projecteuler.net/problem=13
+=#
+
+using DoubleFloats
+# using JSON
+
+function Problem13()
+    #=
+    Work out the first ten digits of the sum of the following one-hundred
+    50-digit numbers
+    =#
+    return string(sum(parse.(BigInt,readlines("../files/Problem13.txt"))))[1:10]
+end
+
+
+println("Time to evaluate Problem 13:")
+@time Problem13()
+println("")
+println("Result for Problem 13: ", Problem13())
diff --git a/src/Julia/Problems001-050/Problem014.jl b/src/Julia/Problems001-050/Problem014.jl
new file mode 100644
index 0000000..cc9f6a5
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem014.jl
@@ -0,0 +1,59 @@
+#=
+Created on 24 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 14 of Project Euler
+https://projecteuler.net/problem=14
+=#
+
+function chain_length(n)#, terms)
+    length = 0
+    while n > 1
+        n = iseven(n) ? n >> 1 : 3n + 1
+        length += 1
+    end
+    return length
+end
+
+function Problem14()
+    #=
+    The following iterative sequence is defined for the set of positive
+    integers:
+
+    n → n/2 (n is even)
+    n → 3n + 1 (n is odd)
+
+    Using the rule above and starting with 13, we generate the following
+    sequence:
+
+    13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
+
+    It can be seen that this sequence (starting at 13 and finishing at 1)
+    contains 10 terms. Although it has not been proved yet (Collatz Problem),
+    it is thought that all starting numbers finish at 1.
+
+    Which starting number, under one million, produces the longest chain?
+
+    NOTE: Once the chain starts the terms are allowed to go above one million.
+    =#
+
+    ans = 0
+    limit = 1_000_000
+    score = 0
+    for i in 1:2:limit  # no need to check even numbers
+        longest = chain_length(i)
+        if longest > score
+            score = longest
+            ans = i
+        end
+    end
+    return ans
+end
+
+
+println("Time to evaluate Problem 14:")
+@time Problem14()
+println("")
+println("Result for Problem 14: ", Problem14())
diff --git a/src/Julia/Problems001-050/Problem015.jl b/src/Julia/Problems001-050/Problem015.jl
new file mode 100644
index 0000000..d556a9d
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem015.jl
@@ -0,0 +1,28 @@
+using Base: Integer
+#=
+Created on 25 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 15 of Project Euler
+https://projecteuler.net/problem=15
+=#
+
+function Problem15()
+    #=
+    Starting in the top left corner of a 2×2 grid, and only being able to
+    move to the right and down, there are exactly 6 routes to the bottom
+    right corner.
+
+    How many such routes are there through a 20×20 grid?
+    =#
+    n = 20
+    return Integer(factorial(big(2n)) / (factorial(big(n)) * factorial(big(2n - n))))
+end
+
+
+println("Time to evaluate Problem 15:")
+@time Problem15()
+println("")
+println("Result for Problem 15: ", Problem15())
diff --git a/src/Julia/Problems001-050/Problem016.jl b/src/Julia/Problems001-050/Problem016.jl
new file mode 100644
index 0000000..670ae1f
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem016.jl
@@ -0,0 +1,26 @@
+using Base: Integer
+#=
+Created on 26 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 16 of Project Euler
+https://projecteuler.net/problem=16
+=#
+
+function Problem16()
+    #=
+    2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
+
+    What is the sum of the digits of the number 2^1000?
+    =#
+    n = 1000
+    return sum(parse(Int, d) for d in string(2^BigInt(n)))
+end
+
+
+println("Time to evaluate Problem 16:")
+@time Problem16()
+println("")
+println("Result for Problem 16: ", Problem16())
diff --git a/src/Julia/Problems001-050/Problem017.jl b/src/Julia/Problems001-050/Problem017.jl
new file mode 100644
index 0000000..c07a708
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem017.jl
@@ -0,0 +1,66 @@
+#=
+Created on 28 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 17 of Project Euler
+https://projecteuler.net/problem=17
+=#
+
+function num2letters(num, dic)
+    if num <= 20
+        return length(dic[num])
+    elseif num < 100
+        tens, units = divrem(num, 10)
+        return length(dic[tens * 10]) + num2letters(units, dic)
+    elseif num < 1000
+        hundreds, rest = divrem(num, 100)
+        if rest != 0
+            return num2letters(hundreds, dic) + length(dic[100])
+            + length("and") + num2letters(rest, dic)
+        else
+            return num2letters(hundreds, dic) + length(dic[100])
+        end
+    else
+        thousands, rest = divrem(num, 1000)
+        return num2letters(thousands, dic) + length(dic[1000])
+    end
+end
+
+function Problem17()
+    #=
+    If the numbers 1 to 5 are written out in words: one, two, three, four,
+    five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
+
+    If all the numbers from 1 to 1000 (one thousand) inclusive were written
+    out in words, how many letters would be used?
+
+    NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
+    forty-two) contains 23 letters and 115 (one hundred and fifteen) contains
+    20 letters. The use of "and" when writing out numbers is in compliance
+    with British usage.
+    =#
+    nums = Dict(
+        0 => "", 1 => "one", 2 => "two", 3 => "three", 4 => "four", 5 => "five",
+        6 => "six", 7 => "seven", 8 => "eight", 9 => "nine", 10 => "ten",
+        11 => "eleven", 12 => "twelve", 13 => "thirteen", 14 => "fourteen",
+        15 => "fifteen", 16 => "sixteen", 17 => "seventeen", 18 => "eighteen",
+        19 => "nineteen", 20 => "twenty", 30 => "thirty", 40 => "forty",
+        50 => "fifty", 60 => "sixty", 70 => "seventy", 80 => "eighty",
+        90 => "ninety", 100 => "hundred", 1000 => "thousand"
+    )
+
+    n = 1000
+    letters = 0
+    for num in 1:n
+        letters += num2letters(num, nums)
+    end
+    return letters
+end
+
+
+println("Time to evaluate Problem 17:")
+@time Problem17()
+println("")
+println("Result for Problem 17: ", Problem17())
diff --git a/src/Julia/Problems001-050/Problem018.jl b/src/Julia/Problems001-050/Problem018.jl
new file mode 100644
index 0000000..b70b447
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem018.jl
@@ -0,0 +1,55 @@
+#= 
+Created on 01 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 18 of Project Euler
+https://projecteuler.net/problem=18 =#
+
+function Problem18()
+    #= 
+    By starting at the top of the triangle below and moving to adjacent 
+    numbers on the row below, the maximum total from top to bottom is 23.
+    
+                                    3
+                                   7 4
+                                  2 4 6
+                                 8 5 9 3 
+    
+    That is, 3 + 7 + 4 + 9 = 23.
+    
+    Find the maximum total from top to bottom of the triangle above =#
+    triangle = [  # Mutable
+                         [75],
+                        [95,64],
+                      [17,47,82],
+                     [18,35,87,10],
+                   [20, 4,82,47,65],
+                  [19, 1,23,75, 3,34],
+                [88, 2,77,73, 7,63,67],
+               [99,65, 4,28, 6,16,70,92],
+             [41,41,26,56,83,40,80,70,33],
+            [41,48,72,33,47,32,37,16,94,29],
+          [53,71,44,65,25,43,91,52,97,51,14],
+         [70,11,33,28,77,73,17,78,39,68,17,57],
+       [91,71,52,38,17,14,91,43,58,50,27,29,48],
+      [63,66, 4,68,89,53,67,30,73,16,69,87,40,31],
+     [4,62,98,27,23, 9,70,98,73,93,38,53,60, 4,23],
+    ]
+    len_triangle = length(triangle)
+
+    for i in len_triangle - 1:-1:1
+        for j in 1:length(triangle[i])
+            triangle[i][j] += max(triangle[i + 1][j], triangle[i + 1][j + 1])
+        end
+    end
+
+    return triangle[1][1]
+end
+
+
+println("Time to evaluate Problem 18:")
+@time Problem18()
+println("")
+println("Result for Problem 18: ", Problem18())
diff --git a/src/Julia/Problems001-050/Problem019.jl b/src/Julia/Problems001-050/Problem019.jl
new file mode 100644
index 0000000..16771d1
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem019.jl
@@ -0,0 +1,43 @@
+#= 
+Created on 02 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 19 of Project Euler
+https://projecteuler.net/problem=19 =#
+
+using Dates
+
+function Problem19()
+    #= 
+    You are given the following information, but you may prefer to do some
+    research for yourself.
+
+    1 Jan 1900 was a Monday.
+    Thirty days has September, April, June and November.
+    All the rest have thirty-one,
+    Saving February alone,
+    Which has twenty-eight, rain or shine.
+    And on leap years, twenty-nine.
+    A leap year occurs on any year evenly divisible by 4, but not on a century
+    unless it is divisible by 400.
+
+    How many Sundays fell on the first of the month during the twentieth
+    century (1 Jan 1901 to 31 Dec 2000)? =#
+    sundays = 0
+    for y in 1901:2000
+        for m in 1:12
+            if Dates.dayofweek(Date(y, m, 1)) == Dates.Sunday
+                sundays += 1
+            end
+        end
+    end
+    return sundays
+end
+
+
+println("Time to evaluate Problem 19:")
+@time Problem19()
+println("")
+println("Result for Problem 19: ", Problem19())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem020.jl b/src/Julia/Problems001-050/Problem020.jl
new file mode 100644
index 0000000..9dbd59e
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem020.jl
@@ -0,0 +1,27 @@
+#= 
+Created on 03 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 20 of Project Euler
+https://projecteuler.net/problem=20 =#
+
+function Problem20()
+    #= 
+    n! means n × (n − 1) × ... × 3 × 2 × 1
+    
+    For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+    and the sum of the digits in the number 10! is:
+    3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
+    
+    Find the sum of the digits in the number 100! =#
+    fact = factorial(big(100))
+    return sum(parse(Int, d) for d in string(fact))
+end
+
+
+println("Time to evaluate Problem 20:")
+@time Problem20()
+println("")
+println("Result for Problem 20: ", Problem20())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem021.jl b/src/Julia/Problems001-050/Problem021.jl
new file mode 100644
index 0000000..ca7f164
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem021.jl
@@ -0,0 +1,55 @@
+#= 
+Created on 05 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 21 of Project Euler
+https://projecteuler.net/problem=21 =#
+
+
+function divisors(n)
+    divisors = Int64[1]
+    m = round(Int, n / 2)
+    for i in 2:m
+        if n % i == 0
+            push!(divisors, i)
+        end
+    end
+    return divisors
+end
+
+function Problem21()
+    #= 
+    Let d(n) be defined as the sum of proper divisors of n (numbers 
+    less than n which divide evenly into n).
+    If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
+    pair and each of a and b are called amicable numbers.
+
+    For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 
+    44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 
+    1, 2, 4, 71 and 142; so d(284) = 220.
+
+    Evaluate the sum of all the amicable numbers under 10000 =#
+    
+    n = 9999
+    s = zeros(Int, n)
+    amicable = Int64[]
+    for i in 2:n
+        s[i] = sum(divisors(i))
+    end
+    
+    for i in 2:n
+        if s[i] <= n && i != s[i] && i == s[s[i]]
+            push!(amicable, i)
+        end
+    end
+
+    return sum(amicable)
+end
+
+
+println("Time to evaluate Problem 21:")
+@time Problem21()
+println("")
+println("Result for Problem 21: ", Problem21())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem022.jl b/src/Julia/Problems001-050/Problem022.jl
new file mode 100644
index 0000000..11e714d
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem022.jl
@@ -0,0 +1,40 @@
+#=
+Created on 08 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 22 of Project Euler
+https://projecteuler.net/problem=22
+=#
+
+using DelimitedFiles
+
+function Problem22()
+    #=
+    Using names.txt, a 46K text file containing over five-thousand first names,
+    begin by sorting it into alphabetical order. Then working out the
+    alphabetical value for each name, multiply this value by its alphabetical
+    position in the list to obtain a name score.
+
+    For example, when the list is sorted into alphabetical order, COLIN, which
+    is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So,
+    COLIN would obtain a score of 938 × 53 = 49714.
+
+    What is the total of all the name scores in the file?
+    =#
+    file = "/datos/Scripts/Gitea/Project_Euler/src/files/Problem22.txt"
+    names = sort(readdlm(file, ',', String)[:])
+
+    result = 0
+    for (idx,name) in enumerate(names)
+        result += sum(Int(c) - 64 for c in name) * idx
+    end
+    return result
+end
+
+
+println("Time to evaluate Problem 22:")
+@time Problem22()
+println("")
+println("Result for Problem 22: ", Problem22())
diff --git a/src/Julia/Problems001-050/Problem023.jl b/src/Julia/Problems001-050/Problem023.jl
new file mode 100644
index 0000000..bd1d940
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem023.jl
@@ -0,0 +1,60 @@
+#=
+Created on 11 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 23 of Project Euler
+https://projecteuler.net/problem=23
+=#
+
+function Problem23()
+    #=
+    A perfect number is a number for which the sum of its proper divisors is
+    exactly equal to the number. For example, the sum of the proper divisors
+    of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect
+    number.
+
+    A number n is called deficient if the sum of its proper divisors is less
+    than n and it is called abundant if this sum exceeds n.
+
+    As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest
+    number that can be written as the sum of two abundant numbers is 24. By
+    mathematical analysis, it can be shown that all integers greater than 28123
+    can be written as the sum of two abundant numbers. However, this upper
+    limit cannot be reduced any further by analysis even though it is known
+    that the greatest number that cannot be expressed as the sum of two
+    abundant numbers is less than this limit.
+
+    Find the sum of all the positive integers which cannot be written as the
+    sum of two abundant numbers.
+    =#
+    LIMIT = 28123
+    divisorsum = zeros(Int32,LIMIT)
+    for i in range(1, LIMIT, step=1)
+        for j in range(2i, LIMIT, step=i)
+            divisorsum[j] += i
+        end
+    end
+    abundantnums = [i for (i, x) in enumerate(divisorsum) if x > i]
+
+    expressible = falses(LIMIT)
+    for i in abundantnums
+        for j in abundantnums
+            if i + j < LIMIT+1
+                expressible[i + j] = true
+            else
+                break
+            end
+        end
+    end
+
+    ans = sum(i for (i, x) in enumerate(expressible) if x == false)
+    return ans
+end
+
+
+println("Time to evaluate Problem 23:")
+@time Problem23()
+println("")
+println("Result for Problem 23: ", Problem23())
diff --git a/src/Julia/Problems001-050/Problem024.jl b/src/Julia/Problems001-050/Problem024.jl
new file mode 100644
index 0000000..a128b83
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem024.jl
@@ -0,0 +1,34 @@
+#=
+Created on 13 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 24 of Project Euler
+https://projecteuler.net/problem=24
+=#
+
+using Combinatorics
+
+function Problem24()
+    #=
+    A permutation is an ordered arrangement of objects. For example, 3124 is
+    one possible permutation of the digits 1, 2, 3 and 4. If all of the
+    permutations are listed numerically or alphabetically, we call it
+    lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
+
+                    012   021   102   120   201   210
+
+    What is the millionth lexicographic permutation of the digits
+    0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
+    =#
+    digits = [0,1,2,3,4,5,6,7,8,9]
+    _permutations = nthperm(digits, 1_000_000)
+    return join(_permutations)
+end
+
+
+println("Time to evaluate Problem 24:")
+@time Problem24()
+println("")
+println("Result for Problem 24: ", Problem24())
diff --git a/src/Julia/Problems001-050/Problem025.jl b/src/Julia/Problems001-050/Problem025.jl
new file mode 100644
index 0000000..0f9d581
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem025.jl
@@ -0,0 +1,52 @@
+#=
+Created on 15 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 25 of Project Euler
+https://projecteuler.net/problem=25
+=#
+
+function Problem25()
+    #=
+    The Fibonacci sequence is defined by the recurrence relation:
+
+        Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
+
+    Hence the first 12 terms will be:
+
+        F1 = 1
+        F2 = 1
+        F3 = 2
+        F4 = 3
+        F5 = 5
+        F6 = 8
+        F7 = 13
+        F8 = 21
+        F9 = 34
+        F10 = 55
+        F11 = 89
+        F12 = 144
+
+    The 12th term, F12, is the first term to contain three digits.
+
+    What is the index of the first term in the Fibonacci sequence to
+    contain 1000 digits?
+    =#
+    a, b = 1, 1
+    index = 2
+
+    while length(digits(b)) < 1000
+        a, b = big(b), big(b+a)
+        index += 1
+    end
+
+    return index
+end
+
+
+println("Time to evaluate Problem 25:")
+@time Problem25()
+println("")
+println("Result for Problem 25: ", Problem25())
diff --git a/src/Julia/Problems001-050/Problem026.jl b/src/Julia/Problems001-050/Problem026.jl
new file mode 100644
index 0000000..160ccc4
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem026.jl
@@ -0,0 +1,53 @@
+#=
+Created on 16 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 26 of Project Euler
+https://projecteuler.net/problem=26
+=#
+
+function Problem26()
+    #=
+    A unit fraction contains 1 in the numerator. The decimal representation
+    of the unit fractions with denominators 2 to 10 are given:
+
+        1/2	= 	0.5
+        1/3	= 	0.(3)
+        1/4	= 	0.25
+        1/5	= 	0.2
+        1/6	= 	0.1(6)
+        1/7	= 	0.(142857)
+        1/8	= 	0.125
+        1/9	= 	0.(1)
+        1/10 = 	0.1
+
+    Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
+    It can be seen that 1/7 has a 6-digit recurring cycle.
+
+    Find the value of d < 1000 for which 1/d contains the longest recurring
+    cycle in its decimal fraction part.
+    =#
+    cycle_length = 0
+    number_d = 0
+    for number in 3:2:999
+        if number % 5 == 0
+            continue
+        end
+        p = 1
+        while (big(10)^p % number) != 1
+            p += 1
+        end
+        if p > cycle_length
+            cycle_length, number_d = p, number
+        end
+    end
+    return number_d
+end
+
+
+println("Time to evaluate Problem 26:")
+@time Problem26()
+println("")
+println("Result for Problem 26: ", Problem26())
diff --git a/src/Julia/Problems001-050/Problem027.jl b/src/Julia/Problems001-050/Problem027.jl
new file mode 100644
index 0000000..35a5aaf
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem027.jl
@@ -0,0 +1,61 @@
+#=
+Created on 19 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 27 of Project Euler
+https://projecteuler.net/problem=27
+=#
+
+using Primes
+
+function Problem27()
+    #=
+    Euler discovered the remarkable quadratic formula:
+
+        n^2 + n + 41
+
+    It turns out that the formula will produce 40 primes for the consecutive
+    integer values 0≤n≤39. However, when n=40, 40^2+40+41=40(40+1)+41 is
+    divisible by 41, and certainly when n=41,41^2+41+41 is clearly divisible
+    by 41.
+
+    The incredible formula n^2−79n+1601 was discovered, which produces 80
+    primes for the consecutive values 0≤n≤79. The product of the coefficients,
+    −79 and 1601, is −126479.
+
+    Considering quadratics of the form:
+
+        n^2 + an + b
+
+    where |a|<1000, |b|≤1000 and |n| is the modulus/absolute value of n
+    e.g. |11|=11 and |−4|=4
+
+    Find the product of the coefficients, a and b, for the quadratic expression
+    that produces the maximum number of primes for consecutive values of n,
+    starting with n=0.
+    =#
+    LIMIT = 1000
+    consecutive_values = 0
+    c = 0
+    for a in -999:LIMIT-1
+        for b in 0:LIMIT
+            n = 0
+            while isprime((n^2) + (a * n) + b)
+                n += 1
+                if n > consecutive_values
+                    consecutive_values = n
+                    c = a * b
+                end
+            end
+        end
+    end
+    return c
+end
+
+
+println("Time to evaluate Problem 27:")
+@time Problem27()
+println("")
+println("Result for Problem 27: ", Problem27())
diff --git a/src/Julia/Problems001-050/Problem028.jl b/src/Julia/Problems001-050/Problem028.jl
new file mode 100644
index 0000000..17c23a7
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem028.jl
@@ -0,0 +1,49 @@
+#=
+Created on 23 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 28 of Project Euler
+https://projecteuler.net/problem=28
+=#
+
+using BenchmarkTools
+
+function Problem28()
+    #=
+    Starting with the number 1 and moving to the right in a clockwise
+    direction a 5 by 5 spiral is formed as follows:
+
+                                21 22 23 24 25
+                                20  7  8  9 10
+                                19  6  1  2 11
+                                18  5  4  3 12
+                                17 16 15 14 13
+
+    It can be verified that the sum of the numbers on the diagonals is 101.
+
+    What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral
+    formed in the same way?
+    =#
+    size = 1001  # Must be odd
+    ans = 1   # Special case for size 1
+    step = 0
+    i, cur = 1, 1
+    while step < size-1
+        step = i * 2
+        for j in 1:4
+            cur += step
+            ans += cur
+        end
+        i += 1
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 28:")
+@btime Problem28()
+println("")
+println("Result for Problem 28: ", Problem28())
diff --git a/src/Julia/Problems001-050/Problem029.jl b/src/Julia/Problems001-050/Problem029.jl
new file mode 100644
index 0000000..94c2881
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem029.jl
@@ -0,0 +1,38 @@
+#=
+Created on 23 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 29 of Project Euler
+https://projecteuler.net/problem=29
+=#
+
+using BenchmarkTools
+
+function Problem29()
+    #=
+    Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
+
+        2^2=4, 2^3=8, 2^4=16, 2^5=32
+        3^2=9, 3^3=27, 3^4=81, 3^5=243
+        4^2=16, 4^3=64, 4^4=256, 4^5=1024
+        5^2=25, 5^3=125, 5^4=625, 5^5=3125
+
+    If they are then placed in numerical order, with any repeats removed, we
+    get the following sequence of 15 distinct terms:
+
+    4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
+
+    How many distinct terms are in the sequence generated by ab for 2≤a≤100
+    and 2≤b≤100?
+    =#
+    terms = Set(big(a)^b for a in 2:100, b in 2:100)
+    return length(terms)
+end
+
+
+println("Time to evaluate Problem 29:")
+@btime Problem29()
+println("")
+println("Result for Problem 29: ", Problem29())
diff --git a/src/Julia/Problems001-050/Problem030.jl b/src/Julia/Problems001-050/Problem030.jl
new file mode 100644
index 0000000..6922e07
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem030.jl
@@ -0,0 +1,47 @@
+#=
+Created on 25 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 30 of Project Euler
+https://projecteuler.net/problem=30
+=#
+
+using BenchmarkTools
+
+function power_digit_sum(pow, n)
+    s = 0
+    while n > 0
+        (n, r) = divrem(n, 10)
+        s += r^pow
+    end
+    return s
+end
+
+function Problem30()
+    #=
+    Surprisingly there are only three numbers that can be written as the sum
+    of fourth powers of their digits:
+
+        1634 = 1^4 + 6^4 + 3^4 + 4^4
+        8208 = 8^4 + 2^4 + 0^4 + 8^4
+        9474 = 9^4 + 4^4 + 7^4 + 4^4
+
+    As 1 = 14 is not a sum it is not included.
+
+    The sum of these numbers is 1634 + 8208 + 9474 = 19316.
+
+    Find the sum of all the numbers that can be written as the sum of fifth
+    powers of their digits.
+    =#
+    ans = sum(i for i in 2:1_000_000 if i == power_digit_sum(5, i))
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 30:")
+@btime Problem30()
+println("")
+println("Result for Problem 30: ", Problem30())
diff --git a/src/Julia/Problems001-050/Problem031.jl b/src/Julia/Problems001-050/Problem031.jl
new file mode 100644
index 0000000..e139fc9
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem031.jl
@@ -0,0 +1,45 @@
+#=
+Created on 27 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 31 of Project Euler
+https://projecteuler.net/problem=31
+=#
+
+using BenchmarkTools
+using IterTools
+
+function Problem31()
+    #=
+    In the United Kingdom the currency is made up of pound (£) and pence (p).
+    There are eight coins in general circulation:
+
+        1p, 2p, 5p, 10p, 20p, 50p, £1 (100p), and £2 (200p).
+
+    It is possible to make £2 in the following way:
+
+        1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
+
+    How many different ways can £2 be made using any number of coins?
+    =#
+    no_ways = 0
+
+    coins = [2, 5, 10, 20, 50, 100]
+
+    bunch_of_coins = product([0:i:200 for i in coins]...)
+    for money in bunch_of_coins
+        if sum(money) <= 200
+            no_ways += 1
+        end
+    end
+    # consider also the case for 200 coins of 1p
+    return no_ways + 1
+end
+
+
+println("Time to evaluate Problem 31:")
+@btime Problem31()
+println("")
+println("Result for Problem 31: ", Problem31())
diff --git a/src/Julia/Problems001-050/Problem032.jl b/src/Julia/Problems001-050/Problem032.jl
new file mode 100644
index 0000000..3f66546
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem032.jl
@@ -0,0 +1,47 @@
+#=
+Created on 30 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 32 of Project Euler
+https://projecteuler.net/problem=32
+=#
+
+using BenchmarkTools
+
+function Problem32()
+    #=
+    We shall say that an n-digit number is pandigital if it makes use of all
+    the digits 1 to n exactly once; for example, the 5-digit number, 15234, is
+    1 through 5 pandigital.
+
+    The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
+    multiplicand, multiplier, and product is 1 through 9 pandigital.
+
+    Find the sum of all products whose multiplicand/multiplier/product identity
+    can be written as a 1 through 9 pandigital.
+    HINT: Some products can be obtained in more than one way so be sure to only
+    include it once in your sum.
+    =#
+
+    ans = Set()
+    pandigital = join(['1', '2', '3', '4', '5', '6', '7', '8', '9'])
+
+    for x in 1:100
+        for y in 100:10_000
+            # product = x * y
+            if join(sort(collect(string(x) * string(y) * string(x * y)))) == pandigital
+                push!(ans, x * y)
+            end
+        end
+    end
+
+    return sum(ans)
+end
+
+
+println("Time to evaluate Problem 32:")
+@btime Problem32()
+println("")
+println("Result for Problem 32: ", Problem32())
diff --git a/src/Julia/Problems001-050/Problem033.jl b/src/Julia/Problems001-050/Problem033.jl
new file mode 100644
index 0000000..dc32286
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem033.jl
@@ -0,0 +1,51 @@
+#=
+Created on 01 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 33 of Project Euler
+https://projecteuler.net/problem=33
+=#
+
+using BenchmarkTools
+
+function Problem33()
+    #=
+    The fraction 49/98 is a curious fraction, as an inexperienced mathematician
+    in attempting to simplify it may incorrectly believe that 49/98 = 4/8,
+    which is correct, is obtained by cancelling the 9s.
+
+    We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
+
+    There are exactly four non-trivial examples of this type of fraction, less
+    than one in value, and containing two digits in the numerator and denominator.
+
+    If the product of these four fractions is given in its lowest common terms,
+    find the value of the denominator.
+    =#
+    numerator = 1
+    denominator = 1
+    for x in 10:99
+        for y in 10:99
+            if x < y
+                if reverse(digits(x))[2] == reverse(digits(y))[1]
+                    if reverse(digits(y))[2] != 0
+                        if reverse(digits(x))[1] / reverse(digits(y))[2] == x / y
+                            numerator *= x
+                            denominator *= y
+                        end
+                    end
+                end
+            end
+        end
+    end
+
+    return Int(denominator/numerator)
+end
+
+
+println("Time to evaluate Problem 33:")
+@btime Problem33()
+println("")
+println("Result for Problem 33: ", Problem33())
diff --git a/src/Julia/Problems001-050/Problem034.jl b/src/Julia/Problems001-050/Problem034.jl
new file mode 100644
index 0000000..d5088e9
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem034.jl
@@ -0,0 +1,42 @@
+#=
+Created on 01 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 34 of Project Euler
+https://projecteuler.net/problem=34
+=#
+
+using BenchmarkTools
+
+function Problem34()
+    #=
+    145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
+
+    Find the sum of all numbers which are equal to the sum of the factorial
+    of their digits.
+
+    Note: As 1! = 1 and 2! = 2 are not sums they are not included.
+    =#
+
+    ans = 0
+
+    for num in 10:2_540_160
+        sum_of_factorial = 0
+        for digit in reverse(digits(num))
+            sum_of_factorial += factorial(digit)
+            if sum_of_factorial == num
+                ans += sum_of_factorial
+            end
+        end
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 34:")
+@btime Problem34()
+println("")
+println("Result for Problem 34: ", Problem34())
diff --git a/src/Julia/Problems001-050/Problem035.jl b/src/Julia/Problems001-050/Problem035.jl
new file mode 100644
index 0000000..8845cba
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem035.jl
@@ -0,0 +1,72 @@
+#=
+Created on 02 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 35 of Project Euler
+https://projecteuler.net/problem=35
+=#
+
+using BenchmarkTools
+using Combinatorics
+using Primes
+
+# function circular_number(n)
+#     result = []
+#     perms = collect(permutations(digits(n), length(n)))
+#     for i in perms
+#         push!(result, parse(Int, join(i)))
+#     end
+#     return result
+# end
+
+function circular_number(n)
+    if n <10
+        return [n]
+    end
+    digs=digits(n)
+    d=length(digs)
+    cyc=zeros(Int,d)
+    x=[10^i for i in 0:d-1]
+    for i in 1:d
+        cyc[i]=sum(x .* digs[vcat(i:d,1:i-1)])
+    end
+    return cyc
+end
+
+function Problem35()
+    #=
+    The number, 197, is called a circular prime because all rotations of the
+    digits: 197, 971, and 719, are themselves prime.
+
+    There are thirteen such primes below 100:
+    2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
+
+    How many circular primes are there below one million?
+    =#
+    circular_primes = []
+    cnt = 0
+    for i in 2:1_000_000
+        if isprime(i)
+            all_primes = true
+            for j in circular_number(i)
+                if !isprime(j)
+                    all_primes = false
+                    break
+                end
+            end
+            if all_primes
+               cnt +=1 #push!(circular_primes, i)
+            end
+        end
+    end
+
+    return cnt #length(circular_primes)
+end
+
+
+println("Time to evaluate Problem 35:")
+@btime Problem35()
+println("")
+println("Result for Problem 35: ", Problem35())
diff --git a/src/Julia/Problems001-050/Problem036.jl b/src/Julia/Problems001-050/Problem036.jl
new file mode 100644
index 0000000..f61165d
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem036.jl
@@ -0,0 +1,38 @@
+#=
+Created on 04 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 36 of Project Euler
+https://projecteuler.net/problem=36
+=#
+
+using BenchmarkTools
+
+function is_palindrome(num)
+    return num == reverse(num)
+end
+
+function Problem36()
+    #=
+    The decimal number, 585 = 1001001001_2 (binary), is palindromic
+    in both bases.
+
+    Find the sum of all numbers, less than one million, which are palindromic
+    in base 10 and base 2.
+    =#
+    ans = 0
+    for n in 1:2:1_000_000
+        if is_palindrome(digits(n,base=10)) && is_palindrome(digits(n,base=2))
+            ans += n
+        end
+    end
+    return ans
+end
+
+
+println("Time to evaluate Problem 36:")
+@btime Problem36()
+println("")
+println("Result for Problem 36: ", Problem36())
diff --git a/src/Julia/Problems001-050/Problem037.jl b/src/Julia/Problems001-050/Problem037.jl
new file mode 100644
index 0000000..da5dae2
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem037.jl
@@ -0,0 +1,53 @@
+#=
+Created on 07 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 37 of Project Euler
+https://projecteuler.net/problem=37
+=#
+
+using BenchmarkTools
+using Primes
+
+function is_truncatable_prime(number)
+    num_str_rev = reverse(digits(number))
+
+    for (idx,num) in enumerate(join(num_str_rev))
+        if !isprime(parse(Int, join(num_str_rev[idx:end]))) ||
+           !isprime(parse(Int, join(num_str_rev[1:end-idx+1])))
+            return false
+        end
+    end
+    return true
+end
+
+function Problem37()
+    """
+    The number 3797 has an interesting property. Being prime itself, it is
+    possible to continuously remove digits from left to right, and remain
+    prime at each stage: 3797, 797, 97, and 7.
+    Similarly we can work from right to left: 3797, 379, 37, and 3.
+
+    Find the sum of the only eleven primes that are both truncatable from left
+    to right and right to left.
+
+    NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
+    """
+    ans = 0
+    # Statement of the problem says this
+    primes_list = primes(11, 1_000_000)
+    for num in primes_list
+        if is_truncatable_prime(num)
+            ans += num
+        end
+    end
+    return ans
+end
+
+
+println("Time to evaluate Problem 37:")
+@btime Problem37()
+println("")
+println("Result for Problem 37: ", Problem37())
diff --git a/src/Julia/Problems001-050/Problem038.jl b/src/Julia/Problems001-050/Problem038.jl
new file mode 100644
index 0000000..13e2c56
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem038.jl
@@ -0,0 +1,56 @@
+#=
+Created on 08 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 38 of Project Euler
+https://projecteuler.net/problem=38
+=#
+
+using BenchmarkTools
+
+function Problem38()
+    #=
+    Take the number 192 and multiply it by each of 1, 2, and 3:
+
+      192 × 1 = 192
+      192 × 2 = 384
+      192 × 3 = 576
+
+    By concatenating each product we get the 1 to 9 pandigital,
+    192384576. We will call 192384576 the concatenated product of
+    192 and (1,2,3)
+
+    The same can be achieved by starting with 9 and multiplying by
+    1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is
+    the concatenated product of 9 and (1,2,3,4,5).
+
+    What is the largest 1 to 9 pandigital 9-digit number that can
+    be formed as the concatenated product of an integer with
+    (1,2, ... , n) where n > 1?
+    =#
+
+    results = []
+    pandigital = join(['1', '2', '3', '4', '5', '6', '7', '8', '9'])
+    # Number must 4 digits (exactly) to be pandigital
+    # if n > 1
+    for i in 1:10_000
+        integer = 1
+        number = ""
+        while length(number) < 9
+            number *= string(integer * i)
+            if join(sort(collect(number))) == pandigital
+                push!(results,number)
+            end
+            integer += 1
+        end
+    end
+    return maximum(results)
+end
+
+
+println("Time to evaluate Problem 38:")
+@btime Problem38()
+println("")
+println("Result for Problem 38: ", Problem38())
diff --git a/src/Julia/Problems001-050/Problem039.jl b/src/Julia/Problems001-050/Problem039.jl
new file mode 100644
index 0000000..177fcda
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem039.jl
@@ -0,0 +1,49 @@
+#=
+Created on 09 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 39 of Project Euler
+https://projecteuler.net/problem=39
+=#
+
+using BenchmarkTools
+
+function Problem39()
+    #=
+    If p is the perimeter of a right angle triangle with integral length sides,
+    {a,b,c}, there are exactly three solutions for p = 120:
+
+        {20,48,52}, {24,45,51}, {30,40,50}
+
+    For which value of p ≤ 1000, is the number of solutions maximised?
+    =#
+    ans, val = 0, 0
+    for p in 2:2:1000
+        sol = 0
+        for a in 1:p
+            for b in a+1:p-2*a
+                c = p - (a + b)
+                if a^2 + b^2 == c^2
+                    sol += 1
+                elseif a^2 + b^2 > c^2
+                    # As we continue our innermost loop, the left side
+                    # gets bigger, right gets smaller, so we're done here
+                    break
+                end
+            end
+        end
+        if sol > ans
+            ans, val = sol, p
+        end
+    end
+
+    return val
+end
+
+
+println("Time to evaluate Problem 39:")
+@btime Problem39()
+println("")
+println("Result for Problem 39: ", Problem39())
diff --git a/src/Julia/Problems001-050/Problem040.jl b/src/Julia/Problems001-050/Problem040.jl
new file mode 100644
index 0000000..b5c0e4a
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem040.jl
@@ -0,0 +1,40 @@
+#=
+Created on 10 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 40 of Project Euler
+https://projecteuler.net/problem=40
+=#
+
+using BenchmarkTools
+
+function Problem40()
+    #=
+    An irrational decimal fraction is created by concatenating the positive integers:
+
+        0.123456789101112131415161718192021...
+
+    It can be seen that the 12th digit of the fractional part is 1.
+
+    If d_n represents the n^th digit of the fractional part, find the value of the following expression.
+
+        d_1 × d_{10} × d_{100} × d_{1_000} × d_{10_000} × d_{100_000} × d_{1_000_000}
+
+    =#
+    ans = 1
+    fraction = join([i for i in 1:1_000_000])
+
+    for i in 0:6
+         ans *= parse(Int, fraction[10^i])
+    end
+
+    return ans
+end
+
+
+println("Time to evaluate Problem 40:")
+@btime Problem40()
+println("")
+println("Result for Problem 40: ", Problem40())
diff --git a/src/Julia/Problems001-050/Problem041.jl b/src/Julia/Problems001-050/Problem041.jl
new file mode 100644
index 0000000..412242e
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem041.jl
@@ -0,0 +1,43 @@
+#=
+Created on 11 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 41 of Project Euler
+https://projecteuler.net/problem=41
+=#
+
+using BenchmarkTools
+using Primes
+
+function is_pandigital(number)
+    number_ = join(sort(digits(number)))
+    check = join([i for i in 1:length(digits(number))])
+    if number_ == check
+        return true
+    end
+    return false
+end
+
+function Problem41()
+    #=
+    We shall say that an n-digit number is pandigital if it makes
+    use of all the digits 1 to n exactly once. For example, 2143 is
+    a 4-digit pandigital and is also prime.
+
+    What is the largest n-digit pandigital prime that exists?
+    =#
+
+    for ans in 7654321:-1:1
+        if is_pandigital(ans) & isprime(ans)
+            return ans
+        end
+    end
+end
+
+
+println("Time to evaluate Problem 41:")
+@btime Problem41()
+println("")
+println("Result for Problem 41: ", Problem41())
diff --git a/src/Julia/Problems001-050/Problem042.jl b/src/Julia/Problems001-050/Problem042.jl
new file mode 100644
index 0000000..b741859
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem042.jl
@@ -0,0 +1,55 @@
+#=
+Created on 12 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 42 of Project Euler
+https://projecteuler.net/problem=42
+=#
+
+using BenchmarkTools
+using DelimitedFiles
+
+function triangle_number(num)
+    return Int(0.5*num*(num+1))
+end
+
+function word_to_value(word)
+    return sum(Int(letter)-64 for letter in word)
+end
+
+function Problem42()
+    #=
+    The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1);
+    so the first ten triangle numbers are:
+
+        1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+
+    By converting each letter in a word to a number corresponding to its alphabetical
+    position and adding these values we form a word value. For example, the word value
+    for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we
+    shall call the word a triangle word.
+
+    Using words.txt, a 16K text file containing nearly two-thousand common English words,
+    how many are triangle words?
+    =#
+
+    triangular_numbers = [triangle_number(n) for n in 1:26]
+    ans = 0
+    file = "/datos/Scripts/Gitea/Project_Euler/src/files/Problem42.txt"
+    words = sort(readdlm(file, ',', String)[:])
+
+    for word in words
+        if word_to_value(word) in triangular_numbers
+            ans += 1
+        end
+    end
+    return ans
+end
+
+
+println("Time to evaluate Problem 42:")
+@btime Problem42()
+println("")
+println("Result for Problem 42: ", Problem42())
diff --git a/src/Julia/Problems001-050/Problem043.jl b/src/Julia/Problems001-050/Problem043.jl
new file mode 100644
index 0000000..110893c
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem043.jl
@@ -0,0 +1,64 @@
+#=
+Created on 13 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 43 of Project Euler
+https://projecteuler.net/problem=43
+=#
+
+using BenchmarkTools
+using Combinatorics
+
+function Problem43()
+    #=
+    The number, 1406357289, is a 0 to 9 pandigital number because
+    it is made up of each of the digits 0 to 9 in some order, but
+    it also has a rather interesting sub-string divisibility property.
+
+    Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this
+    way, we note the following:
+
+        d2d3d4=406 is divisible by 2
+        d3d4d5=063 is divisible by 3
+        d4d5d6=635 is divisible by 5
+        d5d6d7=357 is divisible by 7
+        d6d7d8=572 is divisible by 11
+        d7d8d9=728 is divisible by 13
+        d8d9d10=289 is divisible by 17
+
+    Find the sum of all 0 to 9 pandigital numbers with this property.
+    =#
+    ans = []
+    pandigital = join(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'])
+
+    for n in permutations(pandigital)
+        n_ =join(n)
+        if n_[1] != 0 && join(sort(n)) == pandigital
+            if parse(Int, n_[8:end]) % 17 == 0
+                if parse(Int, n_[7:9]) % 13 == 0
+                    if parse(Int, n_[6:8]) % 11 == 0
+                        if parse(Int, n_[5:7]) % 7 == 0
+                            if parse(Int, n_[4:6]) % 5 == 0
+                                if parse(Int, n_[3:5]) % 3 == 0
+                                    if parse(Int, n_[2:4]) % 2 == 0
+                                        push!(ans, n_)
+                                    end
+                                end
+                            end
+                        end
+                    end
+                end
+            end
+        end
+    end
+
+    return sum(parse(Int, num) for num in ans)
+end
+
+
+println("Time to evaluate Problem 43:")
+@btime Problem43()
+println("")
+println("Result for Problem 43: ", Problem43())
diff --git a/src/Julia/Problems001-050/Problem044.jl b/src/Julia/Problems001-050/Problem044.jl
new file mode 100644
index 0000000..d7a5e88
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem044.jl
@@ -0,0 +1,51 @@
+#=
+Created on 14 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 44 of Project Euler
+https://projecteuler.net/problem=44
+=#
+
+using BenchmarkTools
+using Combinatorics
+
+function pentagonal(n)
+    return Int(n*(3*n-1)/2)
+end
+
+function Problem44()
+    #=
+    Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2.
+    The first ten pentagonal numbers are:
+
+        1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
+
+    It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their
+    difference, 70 − 22 = 48, is not pentagonal.
+
+    Find the pair of pentagonal numbers, Pj and Pk, for which their
+    sum and difference are pentagonal and D = |Pk − Pj| is minimised.
+
+    What is the value of D?
+    =#
+    dif = 0
+    pentagonal_list = [pentagonal(n) for n in 1:2500]
+    pairs = combinations(pentagonal_list, 2)
+    for p in pairs
+        if reduce(+, p) in pentagonal_list && abs(reduce(-, p)) in pentagonal_list
+            dif = (abs(reduce(-,p)))
+            # the first one found would be the smallest
+            break
+        end
+    end
+
+    return dif
+end
+
+
+println("Time to evaluate Problem 44:")
+@btime Problem44()
+println("")
+println("Result for Problem 44: ", Problem44())
diff --git a/src/Julia/Problems001-050/Problem045.jl b/src/Julia/Problems001-050/Problem045.jl
new file mode 100644
index 0000000..50aea45
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem045.jl
@@ -0,0 +1,45 @@
+#=
+Created on 15 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 45 of Project Euler
+https://projecteuler.net/problem=45
+=#
+
+using BenchmarkTools
+
+function pentagonal(n)
+    return Int(n*(3*n-1)/2)
+end
+
+function hexagonal(n)
+    return Int(n*(2*n-1))
+end
+
+function Problem45()
+    #=
+    Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
+        Triangle 	  	Tn=n(n+1)/2 	  	1, 3, 6, 10, 15, ...
+        Pentagonal 	  	Pn=n(3n−1)/2 	  	1, 5, 12, 22, 35, ...
+        Hexagonal 	  	Hn=n(2n−1) 	  	    1, 6, 15, 28, 45, ...
+
+    It can be verified that T285 = P165 = H143 = 40755.
+
+    Find the next triangle number that is also pentagonal and hexagonal.
+    =#
+    pentagonal_list = Set(pentagonal(n) for n in 2:100_000)
+    # all hexagonal numbers are also triangle numbers!
+    hexagonal_list = Set(hexagonal(n) for n in 2:100_000)
+
+    ans = sort!(collect(intersect(hexagonal_list, pentagonal_list)))
+    # First one is already known
+    return ans[2]
+end
+
+
+println("Time to evaluate Problem 45:")
+@btime Problem45()
+println("")
+println("Result for Problem 45: ", Problem45())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem046.jl b/src/Julia/Problems001-050/Problem046.jl
new file mode 100644
index 0000000..27df21e
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem046.jl
@@ -0,0 +1,53 @@
+#=
+Created on 16 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 46 of Project Euler
+https://projecteuler.net/problem=46
+=#
+
+using BenchmarkTools
+using Primes
+
+function is_goldbach(number)
+    for i in number - 1:-1:1
+        if isprime(i) & ((((number - i) / 2)^0.5)%1==0)
+            return true
+        end
+    end
+    return false
+end
+
+function Problem46()
+    #=
+    It was proposed by Christian Goldbach that every odd composite number 
+    can be written as the sum of a prime and twice a square.
+
+    9 = 7 + 2×1^2
+    15 = 7 + 2×2^2
+    21 = 3 + 2×3^2
+    25 = 7 + 2×3^2
+    27 = 19 + 2×2^2
+    33 = 31 + 2×1^2
+
+    It turns out that the conjecture was false.
+
+    What is the smallest odd composite that cannot be written as the sum 
+    of a prime and twice a square?
+    =#
+    ans = 9
+    while true
+        ans += 2
+        if !isprime(ans) & !is_goldbach(ans)
+            return ans
+        end
+    end
+end
+
+
+println("Time to evaluate Problem 46:")
+@btime Problem46()
+println("")
+println("Result for Problem 46: ", Problem46())
\ No newline at end of file
diff --git a/src/Julia/Problems001-050/Problem047.jl b/src/Julia/Problems001-050/Problem047.jl
new file mode 100644
index 0000000..99765ba
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem047.jl
@@ -0,0 +1,65 @@
+#=
+Created on 18 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 47 of Project Euler
+https://projecteuler.net/problem=47
+=#
+
+using BenchmarkTools
+
+function factor(n)
+    ans = []
+    d = 2
+    while d*d <= n
+        if n % d == 0
+            push!(ans,d)
+            n = n ÷ d
+        else
+            d += 1
+        end
+    end
+    if n > 1
+        push!(ans,n)
+    end
+    return ans
+end
+
+function Problem47()
+    #=
+    The first two consecutive numbers to have two distinct prime factors are:
+
+        14 = 2 × 7
+        15 = 3 × 5
+
+    The first three consecutive numbers to have three distinct prime factors are:
+
+        644 = 2² × 7 × 23
+        645 = 3 × 5 × 43
+        646 = 2 × 17 × 19.
+
+    Find the first four consecutive integers to have four distinct prime factors each.
+    What is the first of these numbers?
+    =#
+    ans = []
+
+    for number in 1:1_000_000
+        if length(ans) == 4
+            break
+        elseif length(Set(factor(number))) == 4
+            push!(ans,number)
+        else
+            ans = []
+        end
+    end
+
+    return ans[1]
+end
+
+
+println("Time to evaluate Problem 47:")
+@btime Problem47()
+println("")
+println("Result for Problem 47: ", Problem47())
diff --git a/src/Julia/Problems001-050/Problem048.jl b/src/Julia/Problems001-050/Problem048.jl
new file mode 100644
index 0000000..b23eeb7
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem048.jl
@@ -0,0 +1,27 @@
+#=
+Created on 18 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 48 of Project Euler
+https://projecteuler.net/problem=48
+=#
+
+using BenchmarkTools
+
+function Problem48()
+    #=
+    The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
+
+    Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
+    =#
+    series = sum(big(i)^i for i in 1:1000)
+    return string(series)[end-9:end]
+end
+
+
+println("Time to evaluate Problem 48:")
+@btime Problem48()
+println("")
+println("Result for Problem 48: ", Problem48())
diff --git a/src/Julia/Problems001-050/Problem049.jl b/src/Julia/Problems001-050/Problem049.jl
new file mode 100644
index 0000000..dc4ca17
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem049.jl
@@ -0,0 +1,44 @@
+#=
+Created on 19 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 49 of Project Euler
+https://projecteuler.net/problem=49
+=#
+
+using BenchmarkTools
+using Primes
+
+function Problem49()
+    #=
+    The arithmetic sequence, 1487, 4817, 8147, in which each of the terms
+    increases by 3330, is unusual in two ways:
+    (i) each of the three terms are prime, and,
+    (ii) each of the 4-digit numbers are permutations of one another.
+
+    There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes,
+    exhibiting this property, but there is one other 4-digit increasing sequence.
+
+    What 12-digit number do you form by concatenating the three terms in this sequence?
+    =#
+    ans = []
+    primes_list = primes(1_000, 10_000)
+
+    for number in primes_list
+        if sort(collect(digits(number))) == sort(collect(digits(number+3330))) == sort(collect(digits(number+6660)))
+            if number+3330 in primes_list && number+6660 in primes_list
+                push!(ans, (string(number)*string(number+3300)*string(number+6660)))
+            end
+        end
+    end
+    # return the second one
+    return ans[2]
+end
+
+
+println("Time to evaluate Problem 49:")
+@btime Problem49()
+println("")
+println("Result for Problem 49: ", Problem49())
diff --git a/src/Julia/Problems001-050/Problem050.jl b/src/Julia/Problems001-050/Problem050.jl
new file mode 100644
index 0000000..e890591
--- /dev/null
+++ b/src/Julia/Problems001-050/Problem050.jl
@@ -0,0 +1,54 @@
+#=
+Created on 20 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 50 of Project Euler
+https://projecteuler.net/problem=50
+=#
+
+using BenchmarkTools
+using Primes
+
+function Problem50()
+    #=
+    The prime 41, can be written as the sum of six consecutive primes:
+
+        41 = 2 + 3 + 5 + 7 + 11 + 13
+
+    This is the longest sum of consecutive primes that adds to a prime below one-hundred.
+
+    The longest sum of consecutive primes below one-thousand that adds to a prime, 
+    contains 21 terms, and is equal to 953.
+
+    Which prime, below one-million, can be written as the sum of the most consecutive primes?
+    =#
+
+    ans = 0
+    result = 0
+    prime_list = primes(1_000_000)
+
+    for i in 1:length(prime_list)
+        sum = 0
+        count = 0
+        for j in prime_list[i:end]
+            sum += j
+            count += 1
+            if isprime(sum) && count > result
+                result = count
+                ans = sum
+            end
+            if sum > 1_000_000
+                break
+            end
+        end
+    end
+    return ans
+end
+
+
+println("Time to evaluate Problem 50:")
+@btime Problem50()
+println("")
+println("Result for Problem 50: ", Problem50())
\ No newline at end of file