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 $2013 - 1734 = \boxed{\textbf{(B)}279}$.