diff --git a/ProjectEuler/Problem12.java b/ProjectEuler/Problem12.java
new file mode 100644
index 000000000..89ea3e0ac
--- /dev/null
+++ b/ProjectEuler/Problem12.java
@@ -0,0 +1,66 @@
+package ProjectEuler;
+/**
+ * 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:
+ * <p>
+ * 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+ * <p>
+ * Let us list the factors of the first seven triangle numbers:
+ * <p>
+ * 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.
+ * <p>
+ * What is the value of the first triangle number to have over five hundred divisors?
+ * <p>
+ * link: https://projecteuler.net/problem=12
+ */
+public class Problem12 {
+
+    /**
+     * Driver Code
+     */
+    public static void main(String[] args) {
+        assert solution1(500) == 76576500;
+    }
+
+    /* 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 int solution1(int number) {
+        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 <= number) {
+
+            // 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;
+        }
+        return n;
+    }
+}