The number \( 5^{867} \) is between \(2^{2013}\) and \(2^{2014}\). How many pairs of integers \((m,n)\) are there such that \(m\) is larger than or equal to 1 but smaller than or equal to 2012 and \[5^n < 2^m < 2^{m+2} < 5^{n+1} ?\]

(A) 278

(B) 279

(C) 280

(D) 281

(E) 282

Could someone provide a solution for this? P.S. sorry if the 5^867 looks really small, I don't know why it's like that...

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TopNewestBetween any two powers of 5 there are either \(2\) or \(3\) powers of \(2\) (because \(2^2<5^1<2^3 \)

Consider the intervals \((5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})\)

We want the number of intervals with \(3\) powers of \(2\).

From the given that \(2^{2013}<5^{867}<2^{2014}\) , We know that these \(867\) intervals together have \(2013\) powers of \(2\) . Let \(x\) of them have \(2\) powers of \(2\) and \(y\) of them have \(3\) powers of \(2\). Thus we have the system

\[ \begin{align*}x+y&=867\\ 2x+3y&=2013\end{align*} \]

From which We get \(y=279\Rightarrow\boxed{B}\)

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I did this the same way too, but with a little experimentation, I found that it was also equal to \(\left\lfloor\left|\dfrac{2013\ln\frac{4}{5}}{\ln5}\right|\right\rfloor.\) Why is that? I can get close to something resembling it, but I can't complete the proof.

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Hi Trevor , I Got :

\({\huge{ \left[ 2012\left(1-\frac{2\log 2}{\log 5}\right)\right] = 279 }}\)

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But how did you get that?

And to generalize, can you prove that the number of ordered pairs \((m,n)\) satisfying the conditions with \(m\) having a maximum of \(k\) is or is not equal to this, or equivalent? \[\left\lfloor\left|\dfrac{k\ln\frac{4}{5}}{\ln5}\right|\right\rfloor\]

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Note That \( n\log 5 < m\log 2 < (m+2)\log 2 < (n+1)\log 5 \)

Which Implies That \( \frac{\log 5}{\log 2}n < m <\frac{\log 5}{\log 2}n+1-\frac{ 2\log 2}{\log 5} \)

So We Have To Choose \(m\) Such That We Can Find A Suitable \(n\) .

This interval has length \( 1-\frac{2\log 2}{\log 5} \)

So The Answer Should Be \(\left[ 2012\left(1-\frac{2\log 2}{\log 5}\right)\right] = 279 \)

Where \([x]\) Denotes the Nearest Integer to \(x\) .

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\(1\leq m\leq 2012 \)

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Thanks for the solution

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