How many sets contains 6 or its multiple given the following conditions? 
MyApproach
I created 
@Edit
S1={1,2,3,4,5} ...B)
S2={2,3,4,5,6}
S3={3,4,5,6,7}
S4={4,5,6,7,8}
S5={5,6,7,8,9}
S6={6,7,8,9,10}
S7={7,8,9,10,11}  ....A)
S8={8,9,10,11,12}
From this information I analyzed that these $8$ sets have $6$ sets that have 6 or its multiple.
Thus $80$ sets will have $60$ elements which contain $6$ or its multiple.
Is my Ans right?Please correct me if I am wrong?
 A: Hints:


*

*The sets have $5$ consecutive elements so can at most contain $1$ multiple of $6$.

*For a multiple of $6$ find out of how many of these sets it will be an element.
A: Note that  if  $x  \equiv 0$ mod$(6)$, the  $S_x$ contains a multiple of  $6$.
If  $x \equiv 2$ mod$(6)$, then  $x+4 \equiv 0 $ mod$(6)$, and so  $S_x$ contains a multiple of  $6$.
Similarly you can show that  $S_x$ contains a multiple of $6$, whenever  $x \equiv  3,4,5$ mod$(6)$.
It remains the case  $x \equiv 1$ mod$(6)$. Indeed,  if $x \equiv 1$ mod$(6)$, then  $x+1\equiv 2$ mod$(6)$, $x+2\equiv 3$ mod$(6)$ , $x+3\equiv 4$ mod$(6)$ , and  $x+4\equiv 5$ mod$(6)$  . Hence  $S_x$ doesnt contain any of the multiples of  $6$. 
Thus the number of $S_x$s having no multiple of $6$, is the number of  integers between  $1$ and  $80$  that have remainder one when divided by  $6$. These integers have the form  $6k+1$ with  $k \in \mathbb{N}$. 
Now find  $K$, such tha  $6K+1 \leq  80 $, so  that  $K \leq  13.1$, so  $K=13$. Hence the number  of such sets is 14.  So  $80-14= 66$. $66$ is the answer.
A: You can see that for x between 1 and 6 there are 5 sets which contains 6 or its multiple. and that will be the same for every x between y and y+5.
There are 80 sets so the answer should be  $\lceil 80-(\frac{80}{6}+1) \rceil$ = 66
