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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Top NewestBetween 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 \(y=279\Rightarrow\boxed{B}\) – Gabriel Merces · 3 years, 4 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 \(\left\lfloor\left|\dfrac{2013\ln\frac{4}{5}}{\ln5}\right|\right\rfloor.\) Why is that? I can get close to something resembling it, but I can't complete the proof. – Trevor B. · 3 years, 4 months ago
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Hi Trevor , I Got :
\({\huge{ \left[ 2012\left(1-\frac{2\log 2}{\log 5}\right)\right] = 279 }}\) – Gabriel Merces · 3 years, 4 months ago
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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? \[\left\lfloor\left|\dfrac{k\ln\frac{4}{5}}{\ln5}\right|\right\rfloor\] – Trevor B. · 3 years, 4 months ago
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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 \(x\) . – Gabriel Merces · 3 years, 4 months ago
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Ah, thank your for the full solution, but why \(2012\) and not \(2013?\) The question asks for "smaller than or equal to \(2013\)." – Trevor B. · 3 years, 4 months ago
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Steven Lee Wrote Wrong, In Truth Is :
\(1\leq m\leq 2012 \) – Gabriel Merces · 3 years, 4 months ago
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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. · 3 years, 4 months ago
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You're Welcome ! No Problem ! – Gabriel Merces · 3 years, 4 months ago
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Thanks for the solution – Steven Lee · 3 years, 4 months ago
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