2014 AMC 10B Problem 12

Let $N = 2{,}014{,}000{,}000 = 2014 \cdot 10^6 = (2 \cdot 19 \cdot 53) \cdot 2^6 \cdot 5^6 = 2^7 \cdot 5^6 \cdot 19 \cdot 53$.

Divisors pair up as $d$ and $N/d$, so the $k$-th largest divisor is $N$ divided by the $k$-th smallest. The smallest divisors of $N$, in order, are $1, 2, 4, 5, 8$ (then $10, 16, 19, \dots$); note $3, 6, 7$ do not divide $N$.

The fifth smallest is $8$, so the fifth largest divisor is
$$\frac{N}{8} = \frac{2{,}014{,}000{,}000}{8} = 251{,}750{,}000.$$

The answer is $\boxed{\textbf{(C)}\ 251,750,000}$.

Finding large divisors directly is hopeless, but the pairing $d \leftrightarrow N/d$ turns the question into finding small divisors, which you can list by hand. Factor $N$ first so you know which small numbers actually divide it; the presence of the factor $5$ is what makes the list $1, 2, 4, 5, 8$ rather than pure powers of $2$.

The choices are $N/16$, $N/10$, $N/8$, $N/5$, $N/4$. Ordering the corresponding small divisors $4 < 5 < 8 < 10 < 16$ (with $1$ and $2$ before them) shows $N/8$ is fifth.

• Assuming the small divisors are 1, 2, 4, 8, 16 and answering N/16, forgetting that 5 divides N and slips in before 8.
• Counting $N$ itself as "first largest" but then starting the small list at $2$ instead of $1$, which shifts everything by one.