2014 AMC 10A Problems/Problem 25

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

2014 AMC 10A (Problems • Answer Key • Resources)
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2014 AMC 12A (Problems • Answer Key • Resources)
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2014 AMC 10A Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

Video Solution by Richard Rusczyk

See Also