Difference between revisions of "2014 AMC 10A Problems/Problem 25"

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{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}}
 
{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}}
==ProbIem==
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==Problem==
  
 
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>.  How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath>
 
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>.  How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath>

Revision as of 02:31, 31 October 2016

The following problem is from both the 2014 AMC 12A #22 and 2014 AMC 10A #25, so both problems redirect to this page.

Problem

The number $5^{867}$ is between $2^{2013}$ and $2^{2014}$. How many pairs of integers $(m,n)$ are there such that $1\leq m\leq 2012$ and \[5^n<2^m<2^{m+2}<5^{n+1}?\] $\textbf{(A) }278\qquad \textbf{(B) }279\qquad \textbf{(C) }280\qquad \textbf{(D) }281\qquad \textbf{(E) }282\qquad$

Solution

Between any two consecutive 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 \[x+y=867\]\[2x+3y=2013\] from which we get $y=279$, so the answer is $\boxed{\textbf{(B)}}$.

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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