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

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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
 
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>
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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>.]
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 +
 
 +
To get y=279 we simply realize that 5=<math>2^{2}</math>+1. Thus this means for every 2 x's there is 1 y. Thus this means that y=867/3, or 279.
 +
-Reality Writes
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 14:15, 20 September 2021

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


To get y=279 we simply realize that 5=$2^{2}$+1. Thus this means for every 2 x's there is 1 y. Thus this means that y=867/3, or 279. -Reality Writes

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc10a/379

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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