Call the numbers we are looking for good. A number is good if it is divisible by $7$ but not by $11$, or divisible by $11$ but not by $7$. That's what "divisible by exactly $1$ of $7$ and $11$" means.
Let $a$ be the number of numbers from $1$ to $999$ which are divisible by $7$, and let $b$ be the number of numbers which are divisible by $11$. If we add $a$ and $b$, we will have counted the numbers that are divisible by both $7$ and $11$ twice. However, we should not have counted them at all, they are not good.
So to get the count of the good numbers, we should find $a+b$, and take away twice the number of numbers that are divisible by both $7$ and $11$, like $77$, $154$, and a number of others.
Now the counts are straightforward. Let's find $a$. So we are counting the numbers $7\cdot 1$, $7\cdot 2$, $7\cdot 3$, and so on. What is the biggest $k$ such that $7\cdot k\le 999$? Note that $\frac{999}{7}\approx 142.71$. So the biggest integer $k$ such that $7\cdot k\le 999$ is $142$. It follows that $a=142$.
Finding $b$, and finding the number of numbers $\le 999$ divisible by $77$ is done similarly.
I will present a solution which is entirely made up of a series of completely logical statements.
How many numbers between 1 and 1000 are divisible by 11, well 90 because 90 * 11 = 990 which is the largest multiple of 11 that is less than 1000.
So how many of these numbers are divisible by 3? well every 3rd number within the set of numbers that are divisible by 11 therefore there are 90/3 = 30
So there are 30 numbers that are divisible by 11 and 3.
Out of these numbers how many are divisible by 9? Since 3*3=9, every 3rd multiple in the set of numbers that are divisible by 3 and 11 are divisible by 9.
30/3 = 10
Therefore 10 numbers within the set of 30 numbers are divisible by 9.
30-10 = 20
Which is exactly your answer and your method of solving and with this you are correct.
I hope this was helpful.
Best Answer
You can just add up all the multiples of 7 which are less than 1000. The largest number of such numbers is
$7 \lfloor \dfrac{1000}{7} \rfloor=994$
Then, if you add all the multiples of 7 from 7 to 994,
$7+14+...+994=7(1+2+...\dfrac{994}{7})=7(1+2+...142)=7 \dfrac{(142)(143)}{2}=71071$