A bit shorter solution

[ghost]

cats={} for i in range(1,101): cats[i]=0

for round in range(1,101): # 100 rounds for cat in cats: # 100 cats if cat % round == 0: cats[cat] = abs(cats[cat]-1) # 1 = hat, 0 = no hat

for i in cats: if cats[i]==1: print "Cat with hat: {}".format(i)

[mohammadgoli]

I also just figured that out there is my version of the code :P using list options ;)

python code for catwith HAT :P

haveHat = []

counter = 1

while counter <= 101: for i in range(1, 101): if i % counter == 0: if i in haveHat : haveHat.remove(i) else: haveHat.append(i) counter += 1

print (haveHat)

[chopper7]

Here's my solution. I set up a dict of booleans, ran 100 rounds, then printed the keys that were still True at the end.

# Initially cats 1-100 do not have a hat
cat_hats = {k: False for k in range(1, 101)}

# For every n-th cat: if has hat take it off, if no hat put it on
for n in range(1, 101):
    for cat in cat_hats.keys():
        if cat % n == 0:
            cat_hats[cat] = not (cat_hats[cat])

# Output which cats have hats when you're done
for cat,hat in cat_hats.items():
    if hat:
        print('Cat # {} has a hat!'.format(cat))

[mjhea0]

Updated in v 2.0 of the courses. Coming week of 12/19/2016. https://github.com/realpython/about/blob/master/changelog.csv