From 2f2e94424904011b9c96ec405f3781de0fce4a40 Mon Sep 17 00:00:00 2001
From: Sombit Bose <sombit.bose15@gmail.com>
Date: Wed, 30 Sep 2020 20:23:07 +0530
Subject: [PATCH] Create Problem12.java

Added solution for problem 12 of Project Euler
---
 ProjectEuler/Problem12.java | 63 +++++++++++++++++++++++++++++++++++++
 1 file changed, 63 insertions(+)
 create mode 100644 ProjectEuler/Problem12.java

diff --git a/ProjectEuler/Problem12.java b/ProjectEuler/Problem12.java
new file mode 100644
index 00000000..d85721d7
--- /dev/null
+++ b/ProjectEuler/Problem12.java
@@ -0,0 +1,63 @@
+/*
+    The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
+    The first ten terms would be:
+
+    1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+
+    Let us list the factors of the first seven triangle numbers:
+
+     1: 1
+     3: 1,3
+     6: 1,2,3,6
+    10: 1,2,5,10
+    15: 1,3,5,15
+    21: 1,3,7,21
+    28: 1,2,4,7,14,28
+    We can see that 28 is the first triangle number to have over five divisors.
+
+    What is the value of the first triangle number to have over five hundred divisors?
+*/
+
+public class Problem_12_Highly_Divisible_Triangular_Number {
+
+    /* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
+    public static int triangleNumber(int n) {
+        int sum = 0;
+        for (int i = 0; i <= n; i++)
+            sum += i;
+        return sum;
+    }
+
+    public static void main(String[] args) {
+
+        long start = System.currentTimeMillis(); // start the stopwatch
+        
+        int j = 0; // j represents the jth triangle number
+        int n = 0; // n represents the triangle number corresponding to j
+        int numberOfDivisors = 0; // number of divisors for triangle number n
+        
+        while (numberOfDivisors <= 500) {
+
+            // resets numberOfDivisors because it's now checking a new triangle number
+            // and also sets n to be the next triangle number
+            numberOfDivisors = 0;
+            j++;
+            n = triangleNumber(j);
+            
+            // for every number from 1 to the square root of this triangle number,
+            // count the number of divisors
+            for (int i = 1; i <= Math.sqrt(n); i++)
+                if (n % i == 0)
+                    numberOfDivisors++;
+            
+            // 1 to the square root of the number holds exactly half of the divisors
+            // so multiply it by 2 to include the other corresponding half
+            numberOfDivisors *= 2;
+        }
+        
+        long finish = System.currentTimeMillis(); // stop the stopwatch
+        
+        System.out.println(n);
+        System.out.println("Time taken: " + (finish - start) + " milliseconds");
+    }
+}