Problem source

How many integers greater than or equal to zero and less than a million are not divisible by 2 or 5? What is the average value of these integers?
How many integers greater than or equal to zero and less than 4179 are not divisible by 3 or 7? What is the average value of these integers?

Solution source

There are $1\,000\,000$ numbers between 1 and a million (inclusive). $500\,000$ are divisible by $2$, $200\,000$ are divisible by $5$ and $100\,000$ are divisible by both. Therefore there are:

$1\,000\,000 - 500\,000-200\,000+100\,000 = 400\,000$. 

(Alternatively, the only numbers are those which are $1,3,7,9 \pmod{10}$ so there are $4$ every $10$, or $4 \cdot 100\,000$).

We can sum all these values similarly,

\begin{align*}
S &= \underbrace{\sum_{i=1}^{10^6} i}_{\text{all numbers}}-\underbrace{\sum_{i=1}^{5 \cdot 10^5} 2i}_{\text{all multiples of } 2}-\underbrace{\sum_{i=1}^{2 \cdot 10^5} 5i}_{\text{all multiples of } 5}+\underbrace{\sum_{i=1}^{10^5} 10i}_{\text{all multiples of } 5} \\
&= \frac{10^6 \cdot (10^6 + 1)}{2} - \frac{10^6 \cdot (5\cdot 10^5+1)}{2} -  \frac{10^6 \cdot (2\cdot 10^5+1)}{2} +  \frac{10^6 \cdot (10^5+1)}{2} \\
&= \frac{10^6 (10^5 \cdot (10-5-2+1))))}{2} \\
&= \frac{10^6 \cdot 10^5 \cdot 4}{2} \\
&= 2\cdot 10^{11}
\end{align*}

So the average value is $\frac{2 \cdot 10^{11}}{4 \cdot 10^5} = \frac{10^6}{2} = 500\,000$.

(Alternatively, each value can be paired off eg $999\,999$ with $1$ and so on, leaving averages of $500\,000$).

Note that $4197$ is divisible by $3$ and $7$.

Using the same long we have:

$4179 - \frac{4179}{3} - \frac{4179}{7} + \frac{4179}{21} = 4179 - 1393 - 597 + 199 = 2388$.

The sum will be:

\begin{align*}
S &= \underbrace{\sum_{i=1}^{4179}i }_{\text{all numbers}}- \underbrace{\sum_{i=1}^{1393}3i }_{\text{multiples of }3}- \underbrace{\sum_{i=1}^{597}7i }_{\text{multiples of }7}+ \underbrace{\sum_{i=1}^{199}21i }_{\text{mulitples of }21} \\
&= \frac{4179 \cdot 4180}{2} -  \frac{4179 \cdot 1394}{2} - \frac{4179 \cdot 598}{2} +\frac{4179 \cdot 200}{2} \\
&= \frac{4179 \cdot 2388}{2} 
\end{align*}
So the average value is $\frac{4179}{2}$.