2021 AMC Fall 10B Problem 6

The divisor-counting rule. Take $12 = 2^2 \times 3^1$. A divisor of $12$ picks how many $2$s to use ($0$, $1$, or $2$: three choices) and how many $3$s ($0$ or $1$: two choices). So $12$ has $3 \times 2 = 6$ divisors. In general, add $1$ to each exponent and multiply.

1. What exponents give 2021? Factor $2021 = 43 \times 47$. Both $43$ and $47$ are prime, so $2021$ can only be written as a product of whole numbers bigger than $1$ in two ways: $2021$ by itself, or $43 \times 47$.
- $2021$ by itself: one prime with exponent $2020$, like $2^{2020}$.
- $43 \times 47$: two primes with exponents $42$ and $46$.
2. Make the number small. Use the smallest primes, $2$ and $3$. Put the bigger exponent on the smaller prime, since that costs less:
$$2^{46} \times 3^{42}.$$
This beats $2^{2020}$ easily: $3$ is less than $2^2$, so $3^{42}$ is less than $2^{84}$, and the whole number is less than $2^{46 + 84} = 2^{130}$.
3. Put it in the form $m \times 6^k$. Each pair of one $2$ and one $3$ makes a $6$. There are $42$ threes, so pair them with $42$ of the twos. That leaves $46 - 42 = 4$ twos:
$$2^{46} \times 3^{42} = 2^4 \times (2 \times 3)^{42} = 16 \times 6^{42}.$$
4. Here $m = 16$, which is not divisible by $6$, and $k = 42$. So $m + k = 16 + 42 = 58$.

The answer is $\boxed{\textbf{(B)}\ 58}$.

The number of divisors depends only on the exponents, not on which primes you use. So "smallest number with this many divisors" has two steps: factor the divisor count to find the possible exponents, then give the biggest exponent to $2$, the next to $3$, and so on. Splitting into more small exponents almost always beats one huge exponent.

If $N = p_1^{e_1} p_2^{e_2} \cdots$, the number of divisors is $(e_1 + 1)(e_2 + 1)\cdots$. We need this product to equal $2021 = 43 \cdot 47$.

Since $2021$ has only the factorizations $2021$ and $43 \cdot 47$, either $N = p^{2020}$ or $N = p^{46} q^{42}$ for distinct primes $p, q$. To make $N$ as small as possible, use the smallest primes and give the larger exponent to the smaller prime: the candidates are $2^{2020}$ and $2^{46} \cdot 3^{42}$. The second is far smaller (it is less than $2^{46} \cdot 2^{84} = 2^{130}$).

Now write $2^{46} \cdot 3^{42}$ in the required form:
$$2^{46} \cdot 3^{42} = 2^4 \cdot (2^{42} \cdot 3^{42}) = 16 \cdot 6^{42}.$$
Here $m = 16$ is not divisible by $6$ and $k = 42$, so $m + k = 16 + 42 = 58$.

The answer is $\boxed{\textbf{(B)}\ 58}$.