The problem is as follows:
A clocktower has a mechanism which strikes a bell and this indicates the number of hours which is exactly the same as the number of bell tollings. To indicate $2^n\,a.m$ it uses $(2^n+1)$ seconds; and to indicate its $\textrm{7 a.m}$ it uses $(2^{n+1}+2)$ seconds. If the time between each bell tolling is always the same. What hour will the clocktower indicate in a time of $(4^n-1)$ seconds?. Assume $(n>1)$
The alternatives are as follows:
$\begin{array}{ll} 1.&\textrm{1 a.m}\\ 2.&\textrm{11 a.m}\\ 3.&\textrm{9 a.m}\\ 4.&\textrm{10 a.m}\\ \end{array}$
I'm totally lost on this one. My instructor suggested the use of a formula which it seems kind of obvious and it goes as this:
$\textrm{number of bell tollings}=\frac{\textrm{total time elapsed}}{\textrm{time elapsed between each tolling}}+1$
I'm assuming this is to avoid the flagpole error. Can someone help me with this question?. The sort of answer which I am looking is one which adheres to the formula which has been stated, as well it is the way how I'm supposed to answer this, but I also appreciate an alternate method, since I want to know how to solve this quickly and avoid getting confused. Can someone help me please? By the way, the answer is $\textrm{10 a.m}$ but I don't know how to get there.
Assuming "$2^n+1+2$" (at $7$ a.m.) is a typo for "$2^{n+1}+2$", but that "$2^n+1$" (at $2^n$ a.m.) is not a typo ...
Let $t$ be the time (in seconds) between tollings, and define $m := 2^n$. Noting that $2^{n+1}=2\cdot2^n=2m$ and $4^n = 2^{2n}=m^2$, we have three equations
$$\begin{align} m &= 1 + \frac{m+1}{t} \qquad\to\qquad t(m-1)=m+1 \tag{1}\\ 7 &= 1 + \frac{2m+2}{t} \qquad\to\qquad m = 3t - 1 \tag{2}\\ x &= 1 + \frac{m^2-1}{t} \tag{3} \end{align}$$ Substituting $m$ from $(2)$ into $(1)$ yields $$t(3t-5) = 0 \tag{4}$$ Therefore, $t=0$ (extraneous) or $t=5/3$. In the latter case, $m=4$ (a nice power of $2$), so that, by $(3)$, we have $x = 10$, as expected. $\square$