so it should be 7AM… whatever lets just focus.

You can find if the number is a prime number by using the simple modulo function.

I advise you create a function called `checkIfPrime(number)`

.

First, you’d have to check if the number is greater than 1. So since checkIfPrime() returns a True or a False, you’d check to see if it’s less than or equal to 1. (Use `number <= 1`

), and return False if that’s the case. Otherwise, iterate through all cases using the element n and range 2 and the number. Then use the modulo(%) to check if diving the number with the element n returns 0 or not. (aka. if it’s divisible perfectly by another number that isn’t 1 or itself). If there is a number, then return False, else, that means our number **is** a prime, so return True.

But this means number less than or equal to 1

Mumbo_6

I think it says pretty clearly…

Like this?

```
def checkIfPrime(number):
if number <= 1:
return false
elif number % i == 0:
return true
```

Sorry it took me a while,

Mumbo_6

You forgot the whole step ^^^

sorry, it says “i is not defined”

How do I do that?

Mumbo_6

Remember for loops? Use those.

is this how you mean? Sorry if I make you annoyed by how stupid I am. (I really am )

```
def checkIfPrime(number):
if number <= 1:
return false
for n in range:
return true
```

Mumbo_6

Now theres this error in my way

You’re getting close, but you forgot to specify the **range**. I’ve stated it up here ^^^

Also make sure to **capitalize** `true`

and `false`

for python

if I put `for n in range 2:`

it says I cannot put a number after range. Also, how do I define range?

Mumbo_6

`for n in range(x,y)`

. In this case, you’d use this:

what is the number? it still does not work if I put: `for n in range 2 n:`

The number is the value of your mine.

but how do I write the number of the mine? This function is the first function in my code.

```
def checkIfPrime(number):
if number <= 1:
return False
for n in range 2:
return True
friends = hero.findFriends()
mines = hero.findHazards()
firstMines = []
for mine in mines:
if mine.pos.y > 30 and mine.pos.x < 80:
if checkIfPrime(mine.value):
firstMines.append(mine)
for mine in firstMines:
if mine.pos.x > (hero.pos.x - 25):
hero.moveXY(mine.pos.x + 10, mine.pos.y)
hero.moveXY(mine.pos.x, mine.pos.y)
hero.moveXY(hero.pos.x - 10, hero.pos.y)
for friend in friends:
hero.command(friend, "move", {"x": mine.pos.x + 10, "y": mine.pos.y})
hero.wait(1.6)
hero.command(friend, "move", {"x": mine.pos.x, "y": mine.pos.y})
hero.command(friend, "move", {"x": mine.pos.x - 10, "y": mine.pos.y})
hero.moveXY(2, 15)
for friend in friends:
hero.command(friend, "move", {"x": 2, "y": 15})
```

Mumbo_6

Since `number`

is a parameter in the function `checkIfPrime()`

, you can define that number within the function. For example, you wrote `checkIfPrime(mine.value):`

, so calling `number`

in the function `checkIfPrime`

will use the value of the mine(`mine.value`

)

If you mean this:

```
def checkIfPrime(number):
if number <= 1:
return False
for n in range (mine.value):
return True
```

My hero walks into a non-prime trap.

Mumbo_6