2014 AMC 10A Problem 25

Since $2^2<5<2^3$, each interval $(5^n,5^{n+1})$ contains either two or three powers of $2$. The desired inequality holds exactly for intervals containing three such powers. For $0\le n<867$, let $d$ be the number of intervals with two powers of $2$, and let $t$ be the number with three powers of $2$. Then $d+t=867$. Because $2^{2013}<5^{867}<2^{2014}$, these intervals contain $2013$ powers of $2$ altogether, so $2d+3t=2013$. Solving the system gives $t=279$. Thus, B is the correct answer.