2021 AMC Fall 10B Problem 6

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 $\boxed{\textbf{(B)}\ 58}$.

The divisor-count function depends only on the exponents, so "least integer with $d$ divisors" is a two-step optimization: factor $d$ to determine the possible exponent patterns, then assign larger exponents to smaller primes. Splitting $d$ into more factors almost always beats a single huge exponent.

• Using 2^2020 (a single prime) as the smallest number with 2021 divisors, or writing the answer as 2^4 * 6^42 but then reporting m = 4 or m = 2.
• Assigning exponents the wrong way around, $2^{42} \cdot 3^{46} = 81 \cdot 6^{42}$, which is larger.