Fix error in calculation

src/Julia/Problems001-050/Problem017.jl

@@ -1,4 +1,4 @@
-#=
+#= 
 Created on 28 Jul 2021
 
 @author: David Doblas Jiménez
@@ -9,38 +9,7 @@ https://projecteuler.net/problem=17 =#
 
 using BenchmarkTools
 
-function num2letters(num, dic)
+function num2letters(num)
-    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",
@@ -51,10 +20,43 @@ function Problem17()
         90 => "ninety", 100 => "hundred", 1000 => "thousand"
     )
 
+    if num <= 20
+        return length(nums[num])
+    elseif num < 100
+        tens, units = divrem(num, 10)
+        return (length(nums[Int(tens) * 10]) + num2letters(units))
+    elseif num < 1000
+        hundreds, rest = divrem(num, 100)
+        if rest != 0
+            return (num2letters(hundreds) + length(nums[100])
+            + length("and") + num2letters(rest))
+        else
+            return (num2letters(hundreds) + length(nums[100]))
+        end
+    else
+        thousands, rest = divrem(num, 1000)
+        return (num2letters(thousands) + length(nums[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. =#
+
     n = 1000
     letters = 0
     for num in 1:n
-        letters += num2letters(num, nums)
+        letters += num2letters(num)
     end
     return letters
 end