Write a function that takes an integer and returns an array `[A, B, C]`, where `A` is the number of multiples of 3 (but not 5) below the given integer, `B` is the number of multiples of 5 (but not 3) below the given integer and `C` is the number of multiples of 3 and 5 below the given integer. 

For example, `solution(20)` should return `[5, 2, 1]`

~~~if:r
```r
# in R, returns a numeric vector
solution(20)
[1] 5 2 1

class(solution(20))
[1] "numeric"
```
~~~

Difficulty: EASY
Starter code: def solution(number):
	

Solutions:
def solution(number):
	A = (number - 1) // 3
	B = (number - 1) // 5
	C = (number - 1) // 15
	return [A - C, B - C, C]


def solution(number):
	n = number - 1
	(a, b, c) = (n // 3, n // 5, n // 15)
	return [a - c, b - c, c]


def solution(n):
	n -= 1
	return [n // 3 - n // 15, n // 5 - n // 15, n // 15]


def solution(number):
	number = number - 1
	fizzbuzz = number // 15
	fizz = number // 3 - fizzbuzz
	buzz = number // 5 - fizzbuzz
	return [fizz, buzz, fizzbuzz]


def solution(number):
	[threes, fives, fifteens] = [int((number - 1) / div) for div in [3, 5, 15]]
	return [threes - fifteens, fives - fifteens, fifteens]


def solution(number):
	returnArray = [0, 0, 0]
	for i in range(1, number):
		if i % 15 == 0:
			returnArray[2] += 1
		elif i % 5 == 0 and (not i % 3 == 0):
			returnArray[1] += 1
		elif i % 3 == 0:
			returnArray[0] += 1
	return returnArray


def solution(number):
	number = number - 1
	c = int(number / 15)
	b = int(number / 5) - c
	a = int(number / 3) - c
	return [a, b, c]


def solution(number):
	count = 0
	lst = []
	for i in range(1, number):
		if i % 3 == 0 and (not i % 5 == 0):
			count += 1
	lst.append(count)
	count = 0
	for i in range(1, number):
		if not i % 3 == 0 and i % 5 == 0:
			count += 1
	lst.append(count)
	count = 0
	for i in range(1, number):
		if i % 3 == 0 and i % 5 == 0:
			count += 1
	lst.append(count)
	return lst


def solution(n):
	c = lambda x: (n - 1) // x if x == 15 else (n - 1) // x - c(15)
	return [c(3), c(5), c(15)]