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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]]}}
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==Problem==
 
  
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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>
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<math>\textbf{(A) }278\qquad
 
<math>\textbf{(A) }278\qquad
 
\textbf{(B) }279\qquad
 
\textbf{(B) }279\qquad
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\textbf{(E) }282\qquad</math>
 
\textbf{(E) }282\qquad</math>
  
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==Solution==
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== Solution 1 ==
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Between any two consecutive powers of <math>5</math> there are either <math>2</math> or <math>3</math> powers of <math>2</math> (because <math>2^2<5^1<2^3</math>). Consider the intervals <math>(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})</math>. We want the number of intervals with <math>3</math> powers of <math>2</math>.
  
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Between any two powers of 5 there are either 2 or 3 powers of 2 (because <math>2^2<5^1<2^3</math>). Consider the intervals <math>(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})</math>. We want the number of intervals with 3 powers of 2.
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From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these <math>867</math> intervals together have <math>2013</math> powers of <math>2</math>. Let <math>x</math> of them have <math>2</math> powers of <math>2</math> and <math>y</math> of them have <math>3</math> powers of <math>2</math>. Thus we have the system
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<cmath>x+y=867</cmath><cmath>2x+3y=2013</cmath>
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from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.
  
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From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these 867 intervals together have 2013 powers of 2. Let <math>x</math> of them have 2 powers of 2 and <math>y</math> of them have 3 powers of 2. Thus we have the system
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==Solution 2==
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<cmath>x+y&=867</cmath><cmath>2x+3y&=2013</cmath>
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from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.
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The problem is asking for between how many consecutive powers of <math>5</math> are there <math>3</math> power of <math>2</math>s
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There can be either <math>2</math> or <math>3</math> powers of <math>2</math> between any two consecutive powers of <math>5</math>, <math>5^n</math> and <math>5^{n+1}</math>.
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The first power of <math>2</math> is between <math>5^n</math> and <math>2 \cdot 5^n</math>.
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The second power of <math>2</math> is between <math>2 \cdot 5^n</math> and <math>4 \cdot 5^n</math>.
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The third power of <math>2</math> is between <math>4 \cdot 5^n</math> and <math>8 \cdot 5^n</math>, meaning that it can be between <math>5^n</math> and <math>5^{n+1}</math> or not.
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If there are only <math>2</math> power of <math>2</math>s between every consecutive powers of <math>5</math> up to <math>5^{867}</math>, there would be <math>867\cdot 2 = 1734</math> power of <math>2</math>s. However, there are <math>2013</math> powers of <math>2</math> before  <math>5^{867}</math>, meaning the answer is <math>2013 - 1734 = \boxed{\textbf{(B)}279}</math>.
  
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(Solution by superpi83)
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
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==See Also==
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== Video Solution by Richard Rusczyk ==
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https://www.youtube.com/watch?v=DRJvUMsZtl4&t=4s
  
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== See Also ==
 
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}}
 
{{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}}
 
{{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:51, 15 August 2023

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 1

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 $\boxed{\textbf{(B)}}$.

Solution 2

The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$s

There can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$, $5^n$ and $5^{n+1}$.

The first power of $2$ is between $5^n$ and $2 \cdot 5^n$.

The second power of $2$ is between $2 \cdot 5^n$ and $4 \cdot 5^n$.

The third power of $2$ is between $4 \cdot 5^n$ and $8 \cdot 5^n$, meaning that it can be between $5^n$ and $5^{n+1}$ or not.

If there are only $2$ power of $2$s between every consecutive powers of $5$ up to $5^{867}$, there would be $867\cdot 2 = 1734$ power of $2$s. However, there are $2013$ powers of $2$ before $5^{867}$, meaning the answer is $2013 - 1734 = \boxed{\textbf{(B)}279}$.

~isabelchen

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=DRJvUMsZtl4&t=4s

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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