AMC 2014 question

The number 5867 5^{867} is between 220132^{2013} and 220142^{2014}. How many pairs of integers (m,n)(m,n) are there such that mm is larger than or equal to 1 but smaller than or equal to 2012 and 5n<2m<2m+2<5n+1?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, 5 months ago

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Between any two powers of 5 there are either 22 or 33 powers of 22 (because 22<51<232^2<5^1<2^3

Consider the intervals (50,51),(51,52),(5866,5867)(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})

We want the number of intervals with 33 powers of 22.

From the given that 22013<5867<220142^{2013}<5^{867}<2^{2014} , We know that these 867867 intervals together have 20132013 powers of 22 . Let xx of them have 22 powers of 22 and yy of them have 33 powers of 22. Thus we have the system

x+y=8672x+3y=2013 \begin{aligned}x+y&=867\\ 2x+3y&=2013\end{aligned}

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

Gabriel Merces - 5 years, 5 months ago

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Thanks for the solution

Steven Lee - 5 years, 5 months ago

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I did this the same way too, but with a little experimentation, I found that it was also equal to 2013ln45ln5.\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.

Trevor B. - 5 years, 5 months ago

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Hi Trevor , I Got :

[2012(12log2log5)]=279{\huge{ \left[ 2012\left(1-\frac{2\log 2}{\log 5}\right)\right] = 279 }}

Gabriel Merces - 5 years, 5 months ago

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@Gabriel Merces Those would be equal if you said 2013(12ln2ln5).\left\lfloor2013\left(1-\dfrac{2\ln2}{\ln5}\right)\right\rfloor. Yours needs to say 20132013 instead of 2012.2012. This makes sense, because 58675^{867} becomes part of the ordered pair when the maximum of mm is increased to 2013,2013, but not if the maximum is 2012.2012. Your value is 278.95753278.95753\ldots

But how did you get that?

And to generalize, can you prove that the number of ordered pairs (m,n)(m,n) satisfying the conditions with mm having a maximum of kk is or is not equal to this, or equivalent? kln45ln5\left\lfloor\left|\dfrac{k\ln\frac{4}{5}}{\ln5}\right|\right\rfloor

Trevor B. - 5 years, 5 months ago

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@Trevor B. I Know It's The Same Thing ( I Prefer Use Log )

Note That nlog5<mlog2<(m+2)log2<(n+1)log5 n\log 5 < m\log 2 < (m+2)\log 2 < (n+1)\log 5

Which Implies That log5log2n<m<log5log2n+12log2log5 \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 mm Such That We Can Find A Suitable nn .

This interval has length 12log2log5 1-\frac{2\log 2}{\log 5}

So The Answer Should Be [2012(12log2log5)]=279\left[ 2012\left(1-\frac{2\log 2}{\log 5}\right)\right] = 279

Where [x][x] Denotes the Nearest Integer to xx .

Gabriel Merces - 5 years, 5 months ago

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@Gabriel Merces Ah, thank your for the full solution, but why 20122012 and not 2013?2013? The question asks for "smaller than or equal to 20132013."

Trevor B. - 5 years, 5 months ago

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@Trevor B. Steven Lee Wrote Wrong, In Truth Is :

1m20121\leq m\leq 2012

Gabriel Merces - 5 years, 5 months ago

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@Gabriel Merces 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!

Trevor B. - 5 years, 5 months ago

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@Trevor B. You're Welcome ! No Problem !

Gabriel Merces - 5 years, 5 months ago

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