# AMC 2014 question

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... Note by Steven Lee
5 years, 10 months ago

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Between 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{aligned}x+y&=867\\ 2x+3y&=2013\end{aligned}

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

- 5 years, 10 months ago

Thanks for the solution

- 5 years, 10 months ago

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.

- 5 years, 10 months ago

Hi Trevor , I Got :

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

- 5 years, 10 months ago

Those would be equal if you said $\left\lfloor2013\left(1-\dfrac{2\ln2}{\ln5}\right)\right\rfloor.$ Yours needs to say $2013$ instead of $2012.$ This makes sense, because $5^{867}$ becomes part of the ordered pair when the maximum of $m$ is increased to $2013,$ but not if the maximum is $2012.$ Your value is $278.95753\ldots$

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? $\left\lfloor\left|\dfrac{k\ln\frac{4}{5}}{\ln5}\right|\right\rfloor$

- 5 years, 10 months ago

I Know It's The Same Thing ( I Prefer Use Log )

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 $x$ .

- 5 years, 10 months ago

Ah, thank your for the full solution, but why $2012$ and not $2013?$ The question asks for "smaller than or equal to $2013$."

- 5 years, 10 months ago

Steven Lee Wrote Wrong, In Truth Is :

$1\leq m\leq 2012$

- 5 years, 10 months ago

Oh, I was not aware of that. I don't remember the question, then. Still the same answer, anyways, though, so it's not a problem. Thanks for the help!

- 5 years, 10 months ago

You're Welcome ! No Problem !

- 5 years, 10 months ago