2014 AMC 10B Problem 12

1. Why think about small divisors? Every divisor $d$ of $N$ has a partner $N \div d$. For example, $1$ pairs with $N$ itself, and $2$ pairs with $N \div 2$. The smaller one partner is, the bigger the other. So the largest divisors are $N$ divided by the smallest divisors, in order:
- largest: $N \div 1$,
- second largest: $N \div (\text{second smallest})$,
- and so on.
2. Factor $N$ so we can tell which small numbers divide it:
$$N = 2{,}014{,}000{,}000 = 2014 \times 1{,}000{,}000, \qquad 2014 = 2 \times 19 \times 53, \qquad 1{,}000{,}000 = 2^6 \times 5^6.$$
So $N = 2^7 \times 5^6 \times 19 \times 53$.
3. List the smallest divisors by testing $1, 2, 3, \dots$:
- $1$: yes. $2$: yes. $3$: no (no factor $3$). $4$: yes. $5$: yes. $6$: no. $7$: no. $8$: yes.
So the five smallest divisors are $1, 2, 4, 5, 8$.
4. The fifth largest divisor is therefore
$$N \div 8 = 2{,}014{,}000{,}000 \div 8 = 251{,}750{,}000.$$

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

Finding big divisors directly is hard, but small divisors are easy to list by testing $1, 2, 3, \ldots$ one at a time. Pairing each divisor with its partner turns one job into the other. Factor the number first so you know which small numbers really divide it; here the factor $5$ sneaks in between $4$ and $8$.

Look at the choices: they equal $N \div 16$, $N \div 10$, $N \div 8$, $N \div 5$ and $N \div 4$. The small divisors in order are $1, 2, 4, 5, 8, 10, 16, \ldots$, so $8$ is fifth, and the answer is $N \div 8$.

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