2021 AMC 10B Fall Problem 6

Question

The least positive integer with exactly

2021

distinct positive divisors can be written in the form

m \cdot 6^k,

where

m

and

k

are integers and

6

is not a divisor of

m.

What is

m+k?

47

58

59

88

90

2021=43\cdot47

To minimize the number, put the larger exponent on the smaller prime

Before starting, note that if we can represent the prime factorization of an integer

z

as

z=p_1^{e_1} p_2^{e_2} \cdots,

then there are

(e_1+1)(e_2+1) \cdots

distinct positive factors.

If the number in question has

2021

factors, by the previous logic,

2021=(e_1+1)(e_2+1) \cdots,

and as the prime factorization of

2021 = 43\cdot 47,

then our number must be

p_1^{46}p_2^{42}

or

p^{2020}.

The smallest number we can make in either of these is making

p_1 = 2,p_2=3

in the first configuration, yielding

2^{46}3^{42} = 16\cdot 6^{42} .

Therefore,

m=16

k=42,

so

m+k = 42+16 = 58.

Thus, the answer is B .

Problem 6 in other years