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diff --git a/assignment3.mdwn b/assignment3.mdwn
index 8ea5777c..8c22dfa4 100644
--- a/assignment3.mdwn
+++ b/assignment3.mdwn
@@ -51,19 +51,14 @@ Then `length mylist` evaluates to 3.
 function, write a function that computes factorials.  (Recall that n!,
 the factorial of n, is n times the factorial of n-1.)
 
-Warning: my browser isn't able to compute factorials of numbers
-greater than 2 (it does't provide enough resources for the JavaScript
-interpreter; web pages are not supposed to be that computationally
-intensive).
+	Warning: it takes a long time for my browser to compute factorials larger than 4!
 
-3. (Easy) Write a function `listLenEq` that returns true just in case
-two lists have the
-same length.  That is,
+3. (Easy) Write a function `equal_length` that returns true just in case
+two lists have the same length.  That is,
 
-     listLenEq mylist (make_list junk (make_list junk (make_list junk empty)))
-     ~~> true
+		equal_length mylist (make_list junk (make_list junk (make_list junk empty))) ~~> true
 
-     listLenEq mylist (make_list junk (make_list junk empty))) ~~> false
+		equal_length mylist (make_list junk (make_list junk empty))) ~~> false
 
 
 4. (Still easy) Now write the same function, but don't use the length
@@ -113,7 +108,8 @@ whether the length of the list is less than or equal to 1.  This will
 be your base case for your recursive functions that operate on these
 trees.
 
-1.    Write a function that sums the number of leaves in a tree.
+<OL start=6>
+<LI>Write a function that sums the number of leaves in a tree.
 
 Expected behavior:
 
@@ -136,5 +132,7 @@ Expected behavior:
 	sum-leaves tc ~~> 6
 
 
-2.   Write a function that counts the number of leaves.
+<LI>Write a function that counts the number of leaves.
+
+</OL>