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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.

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  • $\begingroup$ Where you have "$2^n+1+2$", do you mean "$2^{n+1}+2$"? $\endgroup$
    – Blue
    Feb 26, 2020 at 13:33
  • $\begingroup$ @Blue I'm sorry for the delay, yes you spot the error I fixed it right away what it was meant that $2^{n+1}$. $\endgroup$ Mar 15, 2020 at 2:34

1 Answer 1

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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. $\square$

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  • $\begingroup$ Why does it appear three equations?. Wouldn't it be just two as the problem mentions that the clocktower tolls the bell the same number of hours which has elapsed?. One time is $2^n$ and the other is $7\,a.m.$ which would be $7$ times or $7$ tollings. Where does it come the other equation?. What's its justification?. Can you explain better that part?. $\endgroup$ Mar 15, 2020 at 2:37
  • $\begingroup$ @ChrisSteinbeckBell: The first two equations correspond to the two setup facts given in the problem. The third equation corresponds to the actual question: "What hour will the clocktower indicate in $4^n-1$ seconds?" I simply wrote $x$ for the unknown hour. $\endgroup$
    – Blue
    Mar 15, 2020 at 3:29

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