def cube(number):

return "number*number*number"

def by_three(number):

if number%3==0:

return by_three("cube(number)")

```
else:
return False
```

it says cube(1) returned 'number*nmber*number' instead of 1

def cube(number):

return "number*number*number"

def by_three(number):

if number%3==0:

return by_three("cube(number)")

```
else:
return False
```

it says cube(1) returned 'number*nmber*number' instead of 1

Hi, @almightytroll19,

You have this ...

`return "number*number*number"`

With the quotes, `"number*number*number"`

is a string. Remove the quotes, so that it becomes a mathematical expression.

@ryebread4 is correct in that there are additional problems. Removing the quotes, so that the `return`

statement is as follows, will address the original problem that Codecademy reported ...

`return number * number * number`

You could do this instead ...

`return number ** 3`

But, after that correction is made, note that in the `by_three`

function, there are issues in this line that need to be addressed ...

`return by_three("cube(number)")`

The instructions ask that you have it return `cube(number)`

. But, instead, you are calling the `by_three`

function from within itself, and are passing it a string.

why in the second function, by_three is the second line -- if number % 3 == 0? how does this show that the number is divisible by 3? i understand the 0 would indicate that there is no remainder. is that all there is to it? for instance, 12 / 3 equals 4. would the return not be 4?

Hi, @clairethebear88,

Yes, if `number % 3`

yields no remainder, then `number`

is evenly divisible by `3`

. Importantly, the converse is also true. If `number`

is evenly divisible by `3`

, then the result of `number % 3`

is `0`

.

The result of `12 / 3`

is `4`

, as you mentioned, but the result of `12 % 3`

is `0`

. So, you could say that `3`

divides into `12`

exactly `4`

times, and so leaves a remainder of `0`

. However, the result of `13 / 3`

is `4`

and the result of `13 % 3`

is `1`

. In that case, you would say that `3`

divides into `13`

`4`

times, but leaves a remainder of `1`

.