# 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
4 years, 3 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{align*}x+y&=867\\ 2x+3y&=2013\end{align*}

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

- 4 years, 3 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.

- 4 years, 3 months ago

Hi Trevor , I Got :

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

- 4 years, 3 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? $\left\lfloor\left|\dfrac{k\ln\frac{4}{5}}{\ln5}\right|\right\rfloor$

- 4 years, 3 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 $$x$$ .

- 4 years, 3 months ago

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

- 4 years, 3 months ago

Steven Lee Wrote Wrong, In Truth Is :

$$1\leq m\leq 2012$$

- 4 years, 3 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!

- 4 years, 3 months ago

You're Welcome ! No Problem !

- 4 years, 3 months ago

Thanks for the solution

- 4 years, 3 months ago