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

m (→‎Solution)
m (→‎Solution)
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==Solution==
 
==Solution==
  
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Between any two consecutive 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.
+
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>.
  
−
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
+
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
 
<cmath>x+y&=867</cmath><cmath>2x+3y&=2013</cmath>
 
<cmath>x+y&=867</cmath><cmath>2x+3y&=2013</cmath>
 
from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.
 
from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.

Revision as of 14:08, 26 January 2015

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 

\[x+y&=867\] (Error compiling LaTeX. Unknown error_msg)
\[2x+3y&=2013\] (Error compiling LaTeX. Unknown error_msg)

from which we get $y=279$, so the answer is $\boxed{\textbf{(B)}}$.

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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