diff --git a/src/Julia/Problems001-050/Problem001.jl b/src/Julia/Problems001-050/Problem001.jl
index e989b3e..3aaf070 100644
--- a/src/Julia/Problems001-050/Problem001.jl
+++ b/src/Julia/Problems001-050/Problem001.jl
@@ -18,7 +18,7 @@ function Problem1()
     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)
+    ans = sum(x for x in 0:999 if x % 3 == 0 || x % 5 == 0)
 
     return ans
 end
diff --git a/src/Julia/Problems001-050/Problem004.jl b/src/Julia/Problems001-050/Problem004.jl
index c61a06b..4feb1bf 100644
--- a/src/Julia/Problems001-050/Problem004.jl
+++ b/src/Julia/Problems001-050/Problem004.jl
@@ -8,18 +8,14 @@ Solution for Problem 4 of Project Euler
 https://projecteuler.net/problem=4 =#
 
 using BenchmarkTools
-using Profile
 
-function ispalindrome(n::Int64)
-    # return digits(n) == reverse!(digits(n))
-    for i in zip(digits(n), reverse!(digits(n)))
-        if i[1] ≠ i[end]
-            return false
-        end
-    end
-    return true
+
+function ispalindrome(n)
+    s = string(n)
+    s == reverse(s)
 end
 
+
 function Problem4()
     #=
     A palindromic number reads the same both ways. The largest palindrome made
@@ -28,10 +24,9 @@ function Problem4()
     Find the largest palindrome made from the product of two 3-digit numbers. =#
 
     ans = 10_001
-    for i in 100:999, j in i:999
-        num = i * j
-        if (num > ans) && ispalindrome(num)
-            ans = num
+    for i = 100:999, j = i:999
+        if ((i * j > ans) && ispalindrome(i * j))
+            ans = i * j
         end
     end
 
diff --git a/src/Julia/Problems001-050/Problem005.jl b/src/Julia/Problems001-050/Problem005.jl
index 515e27a..18dc5cf 100644
--- a/src/Julia/Problems001-050/Problem005.jl
+++ b/src/Julia/Problems001-050/Problem005.jl
@@ -25,7 +25,7 @@ function Problem5()
     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
+    for i = 1:21
         ans *= i ÷ gcd(i, ans)
     end