From 4d53ec2b825fdc56ea221cab5df580824734ea68 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Fri, 7 Oct 2022 21:36:07 +0200
Subject: [PATCH] Linting

---
 src/Julia/Problems001-050/Problem008.jl | 14 +++++++-------
 src/Julia/Problems001-050/Problem009.jl |  4 ++--
 2 files changed, 9 insertions(+), 9 deletions(-)

diff --git a/src/Julia/Problems001-050/Problem008.jl b/src/Julia/Problems001-050/Problem008.jl
index 55020cc..610c3b6 100644
--- a/src/Julia/Problems001-050/Problem008.jl
+++ b/src/Julia/Problems001-050/Problem008.jl
@@ -19,7 +19,7 @@ function Problem8()
     Find the thirteen adjacent digits in the 1000-digit number that have
     the greatest product. What is the value of this product? =#
 
-    NUM = """
+    NUM::String = """
         73167176531330624919225119674426574742355349194934
         96983520312774506326239578318016984801869478851843
         85861560789112949495459501737958331952853208805511
@@ -42,14 +42,14 @@ function Problem8()
         71636269561882670428252483600823257530420752963450
         """
 
-    num = replace(NUM, "\n" => "")
-    adjacent_digits = 13
-    target = length(num) - adjacent_digits
-    ans, i = 0, 1
+    num::String = replace(NUM, "\n" => "")
+    adjacent_digits::Int16 = 13
+    target::Int32 = length(num) - adjacent_digits
+    ans::Int64, i::Int64 = 0, 1
     while i < target
-        a, j = 1, i
+        a::Int64, j::Int64 = 1, i
         while j < (i + adjacent_digits)
-            a = a * parse(Int, (num[j]))
+            a *= parse(Int, (num[j]))
             j += 1
         end
         if a > ans
diff --git a/src/Julia/Problems001-050/Problem009.jl b/src/Julia/Problems001-050/Problem009.jl
index 865107c..81c7942 100644
--- a/src/Julia/Problems001-050/Problem009.jl
+++ b/src/Julia/Problems001-050/Problem009.jl
@@ -20,8 +20,8 @@ function Problem9()
     Find the product abc. =#
 
     upper_limit = 1000
-    for a in 1:upper_limit + 1
-        for b in a + 1:upper_limit + 1
+    for a = 1:upper_limit+1
+        for b = 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