How many integers $1, 2,....., 11000$ are invertible modulo $880$?
$880$ can be rewritten as $2^4\cdot5\cdot11$.
So I am supposed to find the number of integers in this range that have $2$, $5$ or $11$ as a divisor and then subtract that value from $11000$.
So If I divide $11000$ by each of $2$, $5$ and $11$, I get cardinalities of $5500$, $2200$, and $1000$ respectively. But how exactly am I supposed to find how many integers there are that have both $2$ and $5$ as a divisor, $2$ and $11$ as a divisor, and $5$ and $11$ as a divisor? How am I supposed to find the amount of integers that have all three numbers as a divisor?
Any help?
The inclusion-exclusion principle states that $$\#(A\cup B)=\#(A)+\#(B)-\#(A\cap B)$$
Generalizing this, we also find that $$\#(A\cup B \cup C)=\#(A)+\#(B)+\#(C)-\#(A\cap B) - \#(A\cap C)-\#(B\cap C)+\#(A\cap B\cap C)$$
So, if we let $A,B,C$ be the sets of all integers in $[0,11000]$ divisible by $2$, $5$, and $11$, respectively, the number of invertible elements modulo $880$ is simply the quantity $11000-\#(A\cup B \cup C)$. So, all we need to know is the value of each term in the above formula. But $\#(A\cap B)$ is the number of integers in $[0,11000]$ divisible by $2$ and $5$, i.e. divisible by $2\cdot 5=10$. Similarly, you can find out the meanings of the other terms and plug them into the formula to get your result.