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
No vote yet
1 vote
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
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 \(y=279\Rightarrow\boxed{B}\)
Log in to reply
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.
Log in to reply
Hi Trevor , I Got :
\({\huge{ \left[ 2012\left(1-\frac{2\log 2}{\log 5}\right)\right] = 279 }}\)
Log in to reply
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? \[\left\lfloor\left|\dfrac{k\ln\frac{4}{5}}{\ln5}\right|\right\rfloor\]
Log in to reply
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 \(x\) .
Log in to reply
Ah, thank your for the full solution, but why \(2012\) and not \(2013?\) The question asks for "smaller than or equal to \(2013\)."
Log in to reply
Steven Lee Wrote Wrong, In Truth Is :
\(1\leq m\leq 2012 \)
Log in to reply
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!
Log in to reply
You're Welcome ! No Problem !
Log in to reply
Thanks for the solution
Log in to reply