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

(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...

#NumberTheory

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

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 $y=279\Rightarrow\boxed{B}$

Gabriel Merces - 6 years, 11 months ago

Thanks for the solution

Steven Lee - 6 years, 11 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.

Trevor B. - 6 years, 11 months ago

Hi Trevor , I Got :

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

Gabriel Merces - 6 years, 11 months ago

@Gabriel Merces – 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. - 6 years, 11 months ago

@Trevor B. – 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 - 6 years, 11 months ago

@Gabriel Merces – 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. - 6 years, 11 months ago

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

$1\leq m\leq 2012$

Gabriel Merces - 6 years, 11 months ago

@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. - 6 years, 11 months ago

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

Gabriel Merces - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$