def purify(numbers):

new_num=[]

new_numbers=numbers

for num in numbers:

if num%2!=0:

new_num=new_numbers.remove(num)

return new_num

# 12/15 please help to check

**zeziba**#2

The way you are going about it is kinda wonky.

Let's start with the baisics.

What is the problem?

`Problem: Remove all odd numbers!`

Ok next how do we do that?

```
if number % 2 == 0:
print(even)
```

Ok so we can find our even numbers good. Now we need to collect only the even numbers.

```
even_numbers = []
for number in numbers:
if number % 2 == 0:
even_numbers.append(number)
```

When you use the method remove it will remove the first entry that it finds not all of them.

So instead of removing entries lets just build a list of the items that meet our quilifications.

```
list_numbers = [1, 2, 3, 4, 5, 6]
def purify(number_list):
return_list = []
for number in number_list:
if number % 2 == 0:
return_list.append(number)
return return_list
```

It is harder to remove something then it is to make a new list with what you want.

Good luck!

**eriksvec**#4

You can also do something like this:

```
def purify(l):
for i in l[:]: # l[:] will create a copy of list you are looping through
if i % 2 != 0:
l.remove(i)
return l
```