2014 AMC 10A Problem 25

Interpret the inequality: the three consecutive powers $2^m, 2^{m+1}, 2^{m+2}$ all lie strictly between $5^n$ and $5^{n+1}$. So a pair $(m, n)$ corresponds to a gap $(5^n, 5^{n+1})$ that contains at least three powers of $2$, with $2^m$ the smallest of them.

How many powers of $2$ can such a gap hold? Let $2^m$ be the first power of $2$ above $5^n$, so $2^m < 2 \cdot 5^n$. Then $2^{m+1} < 4 \cdot 5^n < 5^{n+1}$, so every gap holds at least two, and $2^{m+3} > 8 \cdot 5^n > 5^{n+1}$, so no gap holds more than three. Each gap contains exactly $2$ or $3$ powers of $2$, and a gap with $3$ yields exactly one pair.

Now use the given bound. Since $5^{867} > 2^{2013}$ and $5^{867} < 2^{2014}$, the interval $(5^0, 5^{867}) = (1, 5^{867})$ contains exactly the powers $2^1, 2^2, \dots, 2^{2013}$, which is $2013$ powers. This interval is the union of the $867$ gaps $(5^n, 5^{n+1})$ for $n = 0, 1, \dots, 866$, and no power of $2$ equals a power of $5$, so each of those $2013$ powers lies in exactly one gap.

Let $t$ gaps hold three powers and $867 - t$ hold two:
$$3t + 2(867 - t) = 2013 \quad\Longrightarrow\quad t = 2013 - 1734 = 279.$$

Every such gap gives a valid pair: $m \ge 1$, and $m + 2 \le 2013$ means $m \le 2011 \le 2012$. Gaps beyond $5^{867}$ would need $2^m > 5^{867} > 2^{2013}$, violating $m \le 2012$.

The answer is $\boxed{\textbf{(B)}\ 279}$.

Because $4 < 5 < 8$, the powers of $2$ are spaced so that each "power-of-5 gap" swallows either two or three of them; the problem simply asks how many gaps are of the three-type. Counting one set (powers of $2$ up to $2^{2013}$) in two ways, directly and gap by gap, produces a single linear equation. The strange-looking given fact about $5^{867}$ is exactly what fixes the total number of gaps and the total number of powers.

Think of $\log_2 5 \approx 2.3219$. Multiples $n \log_2 5$ for $n = 1, \dots, 866$ chop the interval $(0, 2013)$ into $867$ pieces of length about $2.32$, and a piece contains $3$ integers when its fractional positioning allows. The average piece length $\frac{2013}{867} \approx 2.3218$ means the fraction of three-integer pieces is about $0.32$, and $0.3218 \cdot 867 \approx 279$, confirming the count.

• Counting 2014 powers of 2 (including 2^0 or 2^2014) instead of 2^1 through 2^2013, which gives 280.
• Using $866$ or $868$ gaps instead of $867$ (the gaps are indexed $n = 0$ through $866$), shifting the answer by two.
• Reading the condition as "at least two powers of 2 in the gap" and counting all $867$ gaps, or forgetting that a gap with three powers gives only one pair.