diff --git a/lua/src/p0001.lua b/lua/src/p0001.lua
index 8d32f89e..991838dc 100644
--- a/lua/src/p0001.lua
+++ b/lua/src/p0001.lua
@@ -1,22 +1,38 @@
-return {
-    solution = function()
-        local answer = 0
-
-        for i = 3,999,3
-        do 
-            answer = answer + i
-        end
-
-        for i = 5,999,5
-        do 
-            answer = answer + i
-        end
-
-        for i = 15,999,15
-        do 
-            answer = answer - i
-        end
-
-        return answer
-    end
-}
+-- Project Euler Question 1
+-- 
+-- I did it the old-fashioned way in this language
+-- 
+-- Revision 1:
+-- 
+-- I found a closed form solution that works in the general case, so it's about an
+-- order of magnitude faster now.
+-- 
+-- Problem:
+-- 
+-- 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.
+
+return {
+    solution = function()
+        local answer = 0
+
+        for i = 3,999,3
+        do 
+            answer = answer + i
+        end
+
+        for i = 5,999,5
+        do 
+            answer = answer + i
+        end
+
+        for i = 15,999,15
+        do 
+            answer = answer - i
+        end
+
+        return answer
+    end
+}
diff --git a/lua/src/p0002.lua b/lua/src/p0002.lua
index 7c90b7fe..af559a97 100644
--- a/lua/src/p0002.lua
+++ b/lua/src/p0002.lua
@@ -1,19 +1,34 @@
-return {
-    solution = function()
-        local answer = 0
-        local a = 1
-        local b = 2
-
-        while b < 4000000
-        do
-            answer = answer + b
-        
-            for j = 1,3,1
-            do
-                a, b = b, a + b
-            end
-        end
-
-        return answer
-    end
-}
+-- Project Euler Problem 2
+-- 
+-- I modeled this after the C++ solution, because it was by far the simplest
+-- 
+-- Problem:
+-- 
+-- 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, ...
+-- 
+-- By considering the terms in the Fibonacci sequence whose values do not exceed
+-- four million, find the sum of the even-valued terms.
+-- 
+
+return {
+    solution = function()
+        local answer = 0
+        local a = 1
+        local b = 2
+
+        while b < 4000000
+        do
+            answer = answer + b
+        
+            for j = 1,3,1
+            do
+                a, b = b, a + b
+            end
+        end
+
+        return answer
+    end
+}
diff --git a/lua/src/p0006.lua b/lua/src/p0006.lua
index 1e82c5f1..1ec22bcd 100644
--- a/lua/src/p0006.lua
+++ b/lua/src/p0006.lua
@@ -1,13 +1,31 @@
-return {
-    solution = function()
-        local sum = 1
-        local sum_of_squares = 1
-        for i = 2,100,1
-        do
-            sum = sum + i
-            sum_of_squares = sum_of_squares + (i * i)
-        end
-        return (sum * sum) - sum_of_squares;
-    end
-}
-
+-- Project Euler Problem 6
+-- 
+-- This turned out to be really easy
+-- 
+-- Problem:
+-- 
+-- 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.
+
+return {
+    solution = function()
+        local sum = 1
+        local sum_of_squares = 1
+        for i = 2,100,1
+        do
+            sum = sum + i
+            sum_of_squares = sum_of_squares + (i * i)
+        end
+        return (sum * sum) - sum_of_squares;
+    end
+}
+
diff --git a/lua/src/p0009.lua b/lua/src/p0009.lua
index 0eb09cd4..b4cc8e5e 100644
--- a/lua/src/p0009.lua
+++ b/lua/src/p0009.lua
@@ -1,22 +1,36 @@
-return {
-    solution = function()
-        local c = 3
-        while true
-        do
-            local c_square = c * c
-            for b = 2,c,1
-            do
-                local b_square = b * b
-                for a = 1,b,1
-                do
-                    local a_square = a * a
-                    if a_square + b_square == c_square and a + b + c == 1000
-                    then
-                        return a * b * c
-                    end
-                end
-            end
-            c = c + 1
-        end
-    end
-}
+-- Project Euler Problem 9
+-- 
+-- This was pretty fun to port
+-- 
+-- Problem:
+-- 
+-- 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.
+
+return {
+    solution = function()
+        local c = 3
+        while true
+        do
+            local c_square = c * c
+            for b = 2,c,1
+            do
+                local b_square = b * b
+                for a = 1,b,1
+                do
+                    local a_square = a * a
+                    if a_square + b_square == c_square and a + b + c == 1000
+                    then
+                        return a * b * c
+                    end
+                end
+            end
+            c = c + 1
+        end
+    end
+}
diff --git a/lua/src/p0017.lua b/lua/src/p0017.lua
index ed90d8b6..9180ecb7 100644
--- a/lua/src/p0017.lua
+++ b/lua/src/p0017.lua
@@ -1,75 +1,92 @@
-local function to_string_len(n)
-    if n >= 1000
-    then
-        local thousands = to_string_len(n // 1000 % 100) + 8
-        if n % 1000 ~= 0
-        then
-            thousands = thousands + to_string_len(n % 1000)
-        end
-        return thousands
-    end
-    if n >= 100
-    then
-        local hundreds = to_string_len(n // 100 % 10) + 7
-        if n % 100 ~= 0
-        then
-            hundreds = hundreds + 3 + to_string_len(n % 100)
-        end
-        return hundreds
-    end
-    if n >= 20
-    then
-        local tens_t = {
-            [2] = 6,
-            [3] = 6,
-            [4] = 5,
-            [5] = 5,
-            [6] = 5,
-            [7] = 7,
-            [8] = 6,
-            [9] = 6,
-        }
-        local tens = tens_t[n // 10]
-        if n % 10 ~= 0
-        then
-            tens = tens + to_string_len(n % 10)
-        end
-        return tens
-    end
-    local final = {
-        [0] = 4,
-        [1] = 3,
-        [2] = 3,
-        [3] = 5,
-        [4] = 4,
-        [5] = 4,
-        [6] = 3,
-        [7] = 5,
-        [8] = 5,
-        [9] = 4,
-        [10] = 3,
-        [11] = 6,
-        [12] = 6,
-        [13] = 8,
-        [14] = 8,
-        [15] = 7,
-        [16] = 7,
-        [17] = 9,
-        [18] = 8,
-        [19] = 8,
-    }
-    return final[n]
-end
-
-return {
-    solution = function()
-        local answer = 0
-
-        for x = 1,1000,1
-        do 
-            answer = answer + to_string_len(x)
-        end
-
-        return answer
-    end
-}
+-- Project Euler Problem 17
+-- 
+-- The key here was remembering I don't need to return the whole string, just the length
+-- 
+-- Problem:
+-- 
+-- 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.
+
+local function to_string_len(n)
+    if n >= 1000
+    then
+        local thousands = to_string_len(n // 1000 % 100) + 8
+        if n % 1000 ~= 0
+        then
+            thousands = thousands + to_string_len(n % 1000)
+        end
+        return thousands
+    end
+    if n >= 100
+    then
+        local hundreds = to_string_len(n // 100 % 10) + 7
+        if n % 100 ~= 0
+        then
+            hundreds = hundreds + 3 + to_string_len(n % 100)
+        end
+        return hundreds
+    end
+    if n >= 20
+    then
+        local tens_t = {
+            [2] = 6,
+            [3] = 6,
+            [4] = 5,
+            [5] = 5,
+            [6] = 5,
+            [7] = 7,
+            [8] = 6,
+            [9] = 6,
+        }
+        local tens = tens_t[n // 10]
+        if n % 10 ~= 0
+        then
+            tens = tens + to_string_len(n % 10)
+        end
+        return tens
+    end
+    local final = {
+        [0] = 4,
+        [1] = 3,
+        [2] = 3,
+        [3] = 5,
+        [4] = 4,
+        [5] = 4,
+        [6] = 3,
+        [7] = 5,
+        [8] = 5,
+        [9] = 4,
+        [10] = 3,
+        [11] = 6,
+        [12] = 6,
+        [13] = 8,
+        [14] = 8,
+        [15] = 7,
+        [16] = 7,
+        [17] = 9,
+        [18] = 8,
+        [19] = 8,
+    }
+    return final[n]
+end
+
+return {
+    solution = function()
+        local answer = 0
+
+        for x = 1,1000,1
+        do 
+            answer = answer + to_string_len(x)
+        end
+
+        return answer
+    end
+}
diff --git a/lua/src/p0836.lua b/lua/src/p0836.lua
index e63ee887..901d57d9 100644
--- a/lua/src/p0836.lua
+++ b/lua/src/p0836.lua
@@ -1,5 +1,20 @@
-return {
-    solution = function()
-        return "aprilfoolsjoke"
-    end
-}
+-- Project Euler Problem 836
+-- 
+-- Alright, this was funny
+-- 
+-- Problem:
+-- 
+-- Let A be an **affine plane** over a **radically integral local field** F with residual characteristic p.
+-- 
+-- We consider an **open oriented line section** U of A with normalized Haar measure m.
+-- 
+-- Define f(m,p) as the maximal possible discriminant of the **jacobian** associated to the
+-- **orthogonal kernel embedding** of U into A.
+-- 
+-- Find f(20230401, 57). Give as your answer the concatenation of the first letters of each bolded word.
+
+return {
+    solution = function()
+        return "aprilfoolsjoke"
+    end
+}