Not divisible by $2,3$ or $5$ but divisible by $7$ The question is to determine the number of positive integers up to $2000$ that are not divisible by $2,3$ or $5$ but are divisible by $7$. The answer is supposed to be $76$ but not sure how it was derived
I know that if the question was how many integers are not divisible by $2,3,5$ or $7$ then the answer would be $458$ and I know how to derive this.
 A: A-> divisible by $2$
B -> divisible by $3$
C -> Divisible by $5$
$|A|=|2000/2|= 1000; |B| = |2000/3|= 666; |c| = |2000/5| = 400$
$$n(A \cup B \cup C) = N(A) + N(B) + N(C) - N(A\cap B) - N(A\cap C) - N(B\cap C) + N(A\cap B\cap C)\\
|A\cup B \cup C| = 1000 + 666 + 400 – 333 – 133 – 200 + 66 = 1466$$
No of integers Not Divisible by $2,3$ or $5 = 2000 – 1466 = 534$
Therefore No of integers divisible by $7 = |534/7| = 76$
A: Hint: Given any number $n$ not divisible by $2,3,5$, we get that $7n$ is divisible by $7$ and not divisible by $2,3,5.$ And visa versa.
A: \begin{align}\# \text{ of numbers divisible by }7\text{ but not by }2,3 \text{ or }5 & = \#\text{ of numbers divisible by }7\\
& - \#\text{ of numbers divisible by 7 and 2, i.e., by 14}\\
& - \#\text{ of numbers divisible by 7 and 3, i.e., by 21}\\
& - \#\text{ of numbers divisible by 7 and 5, i.e., by 35}\\
& + \#\text{ of numbers divisible by 7 and 2 and 3, i.e., by 42}\\
& + \#\text{ of numbers divisible by 7 and 2 and 5, i.e., by 70}\\
& + \#\text{ of numbers divisible by 7 and 3 and 5, i.e., by 105}\\
& - \#\text{ of numbers divisible by 7 and 2 and 3 and 5, i.e., by 210}
\end{align}
A: You can solve this with inclusion-exclusion. First look for how many numbers are divisible by $7$: that's $\lfloor 2000/7 \rfloor = 285$, the integer part of $2000/7$.
Now exclude those that are divisible by $2$ (multiples of $14$; there are $\lfloor 2000/14 \rfloor$), divisible by $3$ ($\lfloor 2000/21 \rfloor$) and divisible by $5$ ($\lfloor 2000/35 \rfloor$).
This is excluding too many. Add back those that you removed twice or more: $\lfloor 2000/42 \rfloor + \lfloor 2000/70 \rfloor + \lfloor 2000/105 \rfloor$.
Finally, exclude again those that are divisible by all of $2,3,$ and $5$: these are multiples of $2\cdot 3 \cdot 5 \cdot 7 = 210$ and there are $\lfloor 2000/210 \rfloor$ such.
A: $2000/7= 285+5/7$ so there are $285$  positive multiples of $7$ less than $2000$
Out of every $30$ such multiples $15$ are divisible by $2$, of the remaining $15$, $5$ are divisible by $3$, and of the remaining $10$, $2$ are divisible by $5$ (note the pattern). So this leaves $8$ in every $30$ divisible by $7$ alone.
$285 = 9\cdot 30 +15$. So we have $8\cdot 9=72$ such multiples up to $270$, and it is easy to check $4$ more in the final $15$.
You will spot some patterns here which might lead you to a more efficient general approach. I've suggested "by hand" because it might help you to see why the patterns work - it did the trick for me, many years ago!
A: Let me give you a shortcut:
(1.) first take the LCM of $2,3,5$,and $7$ we will get $210$.
(2.) Now number that will not be divisible by $2$ in $210$ will be $1/2 *210$.--(1)
(3.) numbers divisible by $3$ in 210 is $3\times 70$, $70$ numbers are divisible by $3$ so taking fracton $70/210$ i.e. $1/3$ number are divisible by $3$ i.e. $2/3$ numbers are not divisible by $3$. hence we can write (1) as $210*1/2*2/3$......(2).
(4.) similary for $5$, we have $5*42$, $42$ numbers are divisible by $5$ so $42/210$ means $1/5$ are divisible hence $4/5$ are not divisible hence we can write (2) as   $210*1/2*2/3*4/5$---->(3)
(5.)similary for $7$, $7*30$, $30$ number are divisible in $210$ i.e. $30/210$ convert to $1/7$ hence we can write (3) as $210*1/2*2/3*4/5*1/7$ will give 8.
so in $210$, $8$ numbers are there which will not be divisible by $2,3,5$ and divisible by $7$, so in $2000$ number we can have 
$2000*8/210= 76$ answers. 
