This is a problem that involves adding numbers to items in a list.
In a list you will have to add the item's remainder when divided by a given divisor to each item.

For example if the item is 40 and the divisor is 3 you would have to add 1 since 40 minus the closest multiple of 3 which is 39 is 1. So the 40 in the list will become 41. You would have to return the modified list in this problem.

For this problem you will receive a divisor called `div` as well as simple list of whole numbers called `nums`. Good luck and happy coding.

# Examples
```python
nums = [2, 7, 5, 9, 100, 34, 32, 0], div = 3
  ==>  [4, 8, 7, 9, 101, 35, 34, 0] 

nums = [1000, 999, 998, 997], div = 5
   ==> [1000, 1003, 1001, 999]

nums = [], div = 2
   ==> []
```

**Note:** random tests check lists containing up to 10000 elements.

Difficulty: introductory

Starter code:

def solve(nums,div):
	

Solutions:
def solve(nums,div):
    return [x + x % div for x in nums]



def solve(nums,div): return [n+n%div for n in nums]

solve = lambda n, d: [k + k%d for k in n]


def solve(numbers, divisor):
    return [num + (num % divisor) for num in numbers]


solve = lambda n, d: [i + i % d for i in n]

def solve(nums, div):
    if len(nums) == 0:
        return nums
    
    index = 0
    for num in nums:
        nums[index] += num % div
        index += 1

    return nums


def solve(nums,div):
    for x in range(0, len(nums)):
        nums[x] += nums[x]%div
    return nums

def solve(a, x):
    return [n + n % x for n in a]

def solve(nums,div):
    return [(num%div)+num for num in nums]


def solve(nums,div):
    if nums == []:
        return []
    output = []
    for num in nums:
        output.append((num % div) + num)
    return output