Let X = {5, 6, ..., 200}. How many elements of X are divisible by 2 or by 5?

Step-by-step solution

1. 1

   Choose inclusion–exclusion, and read "or" inclusively. Let $A$ be the multiples of $2$ in $X$ and $B$ the multiples of $5$. "Divisible by $2$ or $5$" means $A\cup B$, and

   $|A\cup B|=|A|+|B|-|A\cap B|$

   The subtraction matters: numbers such as $10$, $20$, $30$ are in both lists and must not be counted twice.

2. 2

   Count the multiples of 2. Use $\big\lfloor \tfrac{200}{2}\big\rfloor-\big\lfloor \tfrac{4}{2}\big\rfloor$ — multiples up to $200$ minus those up to $4$ (the numbers below the start of the range):

   $|A|=100-2=98$

   (The excluded ones are $2$ and $4$; the list runs $6,8,\ldots,200$.)

3. 3

   Count the multiples of 5.

   $|B|=\left\lfloor\frac{200}{5}\right\rfloor-\left\lfloor\frac{4}{5}\right\rfloor=40-0=40$

   (The list is $5,10,\ldots,200$, and $5$ itself is inside the range.)

4. 4

   Count the overlap. A number divisible by both $2$ and $5$ is divisible by $\operatorname{lcm}(2,5)=10$:

   $|A\cap B|=\left\lfloor\frac{200}{10}\right\rfloor-\left\lfloor\frac{4}{10}\right\rfloor=20-0=20$

5. 5

   Apply the formula.

   $|A\cup B|=98+40-20=118$

   $\boxed{118}$

6. 6

   Cross-check with the complement. The set $X$ has $200-5+1=196$ elements. Numbers coprime to $10$ in a block of $10$ consecutive integers number $4$ (those ending in $1,3,7,9$); from $5$ to $200$ there are $19$ full decades $11$–$200$ minus... more simply, the count of numbers ending in $1,3,7,9$ between $5$ and $200$ is $78$ (from $7,9$ then four per decade for the $19$ decades $11$–$200$: $2+4\times 19=78$). Then $196-78=118$, matching.