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

m (minor edit)
m (Solutions)
 
Line 10: Line 10:
 
\textbf{(E) }282\qquad</math>
 
\textbf{(E) }282\qquad</math>
  
== Solutions ==
+
== Solution 1 ==
=== Solution 1 ===
 
 
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>.
 
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>.
  
Line 18: Line 17:
 
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>.
  
=== Video Solution by Richard Rusczyk ===
+
== Video Solution by Richard Rusczyk ==
 
https://artofproblemsolving.com/videos/amc/2014amc10a/379
 
https://artofproblemsolving.com/videos/amc/2014amc10a/379
  

Latest revision as of 22:44, 24 May 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 $\boxed{\textbf{(B)}}$.

Video Solution by Richard Rusczyk

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS