diff --git a/2020/13/input.txt b/2020/13/input.txt
new file mode 100644
index 0000000..66803ac
--- /dev/null
+++ b/2020/13/input.txt
@@ -0,0 +1,2 @@
+1001796
+37,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,41,x,x,x,x,x,x,x,x,x,457,x,x,x,x,x,x,x,x,x,x,x,x,13,17,x,x,x,x,x,x,x,x,23,x,x,x,x,x,29,x,431,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,x,19
\ No newline at end of file
diff --git a/2020/13/part1.js b/2020/13/part1.js
new file mode 100644
index 0000000..828b543
--- /dev/null
+++ b/2020/13/part1.js
@@ -0,0 +1,21 @@
+// this was fairly simple; just keep increasing the wait until you find a bus that leaves. there is probably a more efficient solution but this works
+const fs = require('fs');
+const input = fs.readFileSync('input.txt', 'utf-8').split('\n');
+const time = Number(input[0]);
+const busses = input[1].split(',').filter(v=>v!='x').map(n => Number(n));
+
+let isBus = false;
+let timeDif = 0;
+let busID;
+for (; !isBus; timeDif++) {
+  for(bus of busses) {
+    if ((time + timeDif) % bus == 0){
+      busID = bus;
+      isBus = true;
+      break;
+    }
+  }
+}
+// the for loop does an extra increment at the end oops
+timeDif--
+console.log(timeDif*busID)
\ No newline at end of file
diff --git a/2020/13/part2.js b/2020/13/part2.js
new file mode 100644
index 0000000..ad2c6a6
--- /dev/null
+++ b/2020/13/part2.js
@@ -0,0 +1,19 @@
+// this was a tough one. I still don't fully understand what the puzzle was asking for. but i played around in excel and found this method to find the solution. 
+const fs = require('fs');
+const input = fs.readFileSync('input.txt', 'utf-8').split('\n');
+const busses = input[1].split(',').map(n => Number(n)).map((v,i)=>[v,i]).filter(v => !isNaN(v[0]));
+console.log(busses)
+// start with a timestamp of 0
+let time = 0;
+// if you incrment by the ID of the first bus, the first bus will always meet the pattern described in the puzzle.
+let inc = busses[0][0]
+for (let bus = 1; bus < busses.length; bus++){
+  // continue incrementing until the next bus meets the pattern described in the puzzle
+  while (((time+busses[bus][1]) % busses[bus][0]) != 0) {
+    time += inc
+  }
+  // now you can increment by the new bus that met the pattern, and it will continue to meet the pattern
+  inc*=busses[bus][0]
+}
+// since we looped through all the busses, we know that all busses meet the pattern.
+console.log(time)
\ No newline at end of file
diff --git a/2020/13/test.txt b/2020/13/test.txt
new file mode 100644
index 0000000..e473080
--- /dev/null
+++ b/2020/13/test.txt
@@ -0,0 +1,2 @@
+939
+7,13,x,x,59,x,31,19
\ No newline at end of file
diff --git a/2020/13/test2.txt b/2020/13/test2.txt
new file mode 100644
index 0000000..267507d
--- /dev/null
+++ b/2020/13/test2.txt
@@ -0,0 +1,2 @@
+
+67,7,59,61
\ No newline at end of file
diff --git a/2020/13/test3.txt b/2020/13/test3.txt
new file mode 100644
index 0000000..f96fffd
--- /dev/null
+++ b/2020/13/test3.txt
@@ -0,0 +1,2 @@
+
+67,x,7,59,61
\ No newline at end of file
diff --git a/2020/13/test4.txt b/2020/13/test4.txt
new file mode 100644
index 0000000..7a02b4b
--- /dev/null
+++ b/2020/13/test4.txt
@@ -0,0 +1,2 @@
+
+67,7,x,59,61
\ No newline at end of file
diff --git a/2020/13/test5.txt b/2020/13/test5.txt
new file mode 100644
index 0000000..15576dd
--- /dev/null
+++ b/2020/13/test5.txt
@@ -0,0 +1,2 @@
+
+1789,37,47,1889
\ No newline at end of file
diff --git a/README.md b/README.md
index 0dc85b5..75c4fd9 100644
--- a/README.md
+++ b/README.md
@@ -53,15 +53,4 @@ if (advent) {
 10. [:star: :star:](https://adventofcode.com/2020/day/10 "see puzzle")
 11. [:star: :star:](https://adventofcode.com/2020/day/11 "see puzzle")
 12. [:star: :star:](https://adventofcode.com/2020/day/12 "see puzzle")
-
-## Languages Used
-* Python
-	- py
-* Node JS
- 	- js
-* Java (Processing)
-	- pde
-* HTML, CSS, JavaScript
-	- html
-  - css
-  - js
\ No newline at end of file
+13. [:star: :star:](https://adventofcode.com/2020/day/13 "see puzzle")