From 85b66f8e64576ec3a17066bb1b920a5cd02a30c2 Mon Sep 17 00:00:00 2001
From: Charles Cabergs <me@cacharle.xyz>
Date: Wed, 23 Jun 2021 14:48:30 +0200
Subject: problem 73 in julia

---
 c/012-highly_divisible_triangular_number.c |  2 +-
 julia/073-counting_fractions_in_a_range.jl | 33 ++++++++++++++++++++++++++++++
 2 files changed, 34 insertions(+), 1 deletion(-)
 create mode 100644 julia/073-counting_fractions_in_a_range.jl

diff --git a/c/012-highly_divisible_triangular_number.c b/c/012-highly_divisible_triangular_number.c
index 4ece373..c25f37d 100644
--- a/c/012-highly_divisible_triangular_number.c
+++ b/c/012-highly_divisible_triangular_number.c
@@ -15,7 +15,7 @@ int count_divisors(int nb)
     int i, div_nb = 0;
     for (i = 1; i <= sqrt(nb); i++) {
         if (nb % i == 0)
-            div_nb += (i != nb / i ? 2 : 1);
+            div_nb += (i != nb / i ? 2 : 1);  // how the hell did I get there? eddie woo
     }
     return div_nb;
 }
diff --git a/julia/073-counting_fractions_in_a_range.jl b/julia/073-counting_fractions_in_a_range.jl
new file mode 100644
index 0000000..5ec6c21
--- /dev/null
+++ b/julia/073-counting_fractions_in_a_range.jl
@@ -0,0 +1,33 @@
+###
+# Counting fractions in a range
+# Problem 73
+#
+# Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1,
+# it is called a reduced proper fraction.
+# If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we
+# get:
+# 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5,
+# 5/6, 6/7, 7/8
+# It can be seen that there are 3 fractions between 1/3 and 1/2.
+# How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions
+# for d ≤ 12,000?
+###
+
+
+# HCF means Higest Common Factor (aka Greatest Common Divisor)
+# HCF(n, d) is to ensure that the fraction is reduced.
+# Eddie Woo GCD: https://www.youtube.com/watch?v=ui5yjN1riQA
+
+const TOP = 12_000
+
+# ~15s
+result = length(
+    Set([
+        n // d
+        for d in 1:TOP
+        for n in 1:d
+        if 1 // 3 < n // d < 1 // 2
+    ])
+)
+
+println(result)
-- 
cgit