The book contains $p$ sheets (leafs) and has therefore pagenumbers from $1$ to $2p$.
The sum of all the pagenumbers is then given by
$$
\sum_{i=1}^{2p}i =p \Big( 2p + 1 \Big).
$$
The father holds the page with page number $n$ in his hand, so we need to solve
$$
81,707 = p \Big( 2p + 1 \Big) - n.
$$
As $81,707 \le p \Big( 2p + 1 \Big)$, we obtain
$$
p \ge 202,
$$
but as $n \le 2p$, we obtain
$$
p \Big(2 p + 1 \Big) - 81,707 \le 2 p,
$$
whence
$$
p \le 202,
$$
so the book contains $202$ pages, whence the page number is given by
$$
202 \times 405 - 81,707 = 103.
$$
The question is: if the father is holding a page $x$ does that mean to exclude the pagenumbers on both sides of the page?
Then the page that you father is holding is $51/52$.
Hope you get your book back!
We can translate the statements about Hank and his car thusly:
$$\begin{align}
\text{hank}_{\text{now}} + \text{car}_{\text{now}} &= 56 &(1)\\
\text{hank}_{\text{now}} &= \text{car}_{\text{then}} &(2)\\
\text{car}_{\text{now}} &= 2\cdot \text{hank}_{\text{then}} &(3)
\end{align}$$
And we can use the passage of time to convert from "now" to "then":
$$\begin{align}
\text{hank}_{\text{now}} &=\text{time} + \text{hank}_{\text{then}} \\
\text{car}_{\text{now}} &= \text{time} + \text{car}_{\text{then}}
\end{align}$$
Writing $h := \text{hank}_{\text{then}}$, $c := \text{car}_{\text{then}}$, $t := \text{time}$, we have
$$\begin{align}
(t+h) + (t+c) = 2 t + h + c &= 56 &(1^\prime)\\
t+h &= c &(2^\prime)\\
t + c &= 2h &(3^\prime)
\end{align}$$
Solving gives
$$t = 8 \qquad h = 16 \qquad c = 24$$
So,
$$\text{hank}_{\text{then}} = 16 \qquad \text{hank}_{now} = 16 + 8 = 24$$
$$\text{car}_{\text{then}} = 24 \qquad \text{car}_{now} = 24 + 8 = 32$$
Let's check:
[T]he age of Hank (24), and the age his car (32) combined is 56 years!
[H]is car is (32) twice as old as Hank was (16) when his car was (24) as old as Hank is now (24).
That's it!
Best Answer
Let $x$ be the age of the son and $y$ the father. We have first off $2x=y$ right now. The second sentence says that "when the son is the age of the father...", we note that right now the son is $x$ and his father is $y=2x$, so when the son is the father's age, that is when the son is $2x$ years old, their ages add to 180. The number of years that pass from the time that the son is $x$ till the time the son is $2x$ is simply $x$. Thus $(2x) + (y+x) = 180$. Now we have a system $$2x=y \\ (2x) + (y+x) = 180$$ The solution is $x=36$ and $y=72$. Those are the ages of the two right now, while the ages later on are $72$ and $108$ respectively.