diff --git a/2017/README.md b/2017/README.md
index 1c36982..f3ecd00 100644
--- a/2017/README.md
+++ b/2017/README.md
@@ -16,6 +16,7 @@ The current plan, in no particular order:
 - [ ] Haskell
 - [ ] Java
 - [ ] Kotlin
+- [x] Matlab - [Day 06](./day-06)
 - [ ] Node.js
 - [ ] Objective C
 - [x] Perl - [Day 05](./day-05/solution.pl)
diff --git a/2017/day-06/README.md b/2017/day-06/README.md
new file mode 100644
index 0000000..ae519b2
--- /dev/null
+++ b/2017/day-06/README.md
@@ -0,0 +1,7 @@
+# Usage
+
+Since this solution is in matlab, it works a bit differently. You can
+solve part 1 by calling `solution([original division of wealth])`, which
+returns the number of cycles and the point at which it repeated. You can
+then call `solution` again with that repetition point to get the length
+of that cycle.
diff --git a/2017/day-06/solution.m b/2017/day-06/solution.m
new file mode 100644
index 0000000..d945c99
--- /dev/null
+++ b/2017/day-06/solution.m
@@ -0,0 +1,20 @@
+function [cycles,repeat] = solution (division)
+  division = vec(division);
+  states = zeros(0, size(division)(1));
+  
+  while ! ismember(transpose(division), states, "rows")
+    states = [states; transpose(division)];
+    
+    [maxval, maxindex] = max(division);
+    division(maxindex) = 0;
+    
+    for i = 1:maxval
+      targetIndex = mod(maxindex + i - 1, size(division)(1)) + 1;
+      division(targetIndex) += 1;
+    end
+    
+  end
+  
+  cycles = size(states)(1);
+  repeat = division;
+endfunction