As $$3$$ and $$5$$ are coprime, the squares that have more than one mark are multiples of both $$3$$ and $$5$$, $$\left( {} \right.$$multiples of $$15$$$$\left. {} \right)$$; or multiples of both $$3$$ and $$7$$, $$\left( {} \right.$$multiples of $$21$$$$\left. {} \right)$$; or multiples of $$5$$ and $$7$$, $$\left( {} \right.$$multiples of $$35$$$$\left. {} \right)$$; or multiples of $$3$$, $$5$$ and $$7$$, $$\left( {} \right.$$multiples of $$105$$$$\left. {} \right)$$. However, the latter will be included in all of the first three categories. Between $$1$$ and $$150$$ inclusive, there are ten multiples of $$15$$, seven multiples of $$21$$ and four multiples of $$35$$, making a total of $$21$$ multiples. However, there is one multiple of $$3$$, $$5$$ and $$7$$ between $$1$$ and $$150$$, namely $$105$$. So $$105$$ has been counted three times in those $$21$$ multiples, but corresponds to exactly one marked square. Therefore the total number of marked squares is $$21−2=19$$. 
The positive integers from $$1$$ to $$150$$ inclusive are placed in a $$10$$ by $$15$$ grid so that each cell contains exactly one integer. Then the multiples of $$3$$ are given a red mark, the multiples of $$5$$ are given a blue mark, and the multiples of $$7$$ are given a green mark.  How many cells have more than $$1$$ mark? 
Options:
1: $$10$$ 
2: $$12$$ 
3: $$15$$ 
4: $$18$$ 
5: $$19$$ 