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

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

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