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

(See Also)
m (I changed the phrase "two powers of 5" to "two consecutive powers of 5" for clarity.)
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==Solution==
 
==Solution==
  
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.
+
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.
  
 
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 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

Revision as of 16:31, 14 April 2014

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. ! Misplaced alignment tab character &.)
\[2x+3y&=2013\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)

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

(Solution by superpi83)

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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