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most courses

teachers.py
```# The dictionary will look something like:
# {'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'],
#  'Kenneth Love': ['Python Basics', 'Python Collections']}
#
# Each key will be a Teacher and the value will be a list of courses.
#
# Your code goes below here.

def num_teachers(dic):
return len(dic)

def num_courses(dic):
course = 0
for teachers in dic:
course += len(dic[teachers])
return course

def courses(dic):
class1 = []
for classes in dic.values():
class1.extend(classes)
return class1

def most_courses(dic):
max_count = 0
teachers = " "
for teach, m_courses in dic.iteritems():
if teach > (max_count):
max_count = len(m_courses)
teachers = teach
return teachers
```

I have made a few edits to your code to get it to pass:

1 - Changed .iteritems() to .items(). I think .iteritems() might be a Python 2 thing, but in Python 3, dictionary.items() is the way to go.

2 - Changed your if statement to compare the length of current teacher's courses to the max count. You were comparing the teacher's name to max count.

3 - Indented the 'teachers = teach' line. You only want this variable to change if that teacher is indeed the teacher with the current max count.

Full solution here:

```def most_courses(dic):
max_count = 0
teachers = " "
for teach, m_courses in dic.items():
if len(m_courses) > (max_count):
max_count = len(m_courses)
teachers = teach
return teachers
```

If two or more teachers have same number of courses, the above gives you the last teacher found in the dict. I suppose if you wanted to return a list of teachers you could reset teachers whenever a new max is found, otherwise, if same max append name to a list.

thanks for your help Stuart I really appreciate it.

close, you just need to tab the teachers = teach line over once to the right so it is under max count, not the if, and the challenge doesn't like iteritems so just just use items().