2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 6 Easy

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 ?(2021 AMC Fall 10B, Question #6)

  • A.

    47

  • B.

    58

  • C.

    59

  • D.

    88

  • E.

    90

Answer:B

Solution 1:

Let this positive integer be written as p_{1}^{e_{1}} \cdot p_{2}^{c_{2}}. The number of factors of this number is therefore \left(e_{1}+1\right) \cdot\left(e_{2}+1\right), and this must equal 2021. The prime factorization of 2021 is 43 \cdot 47, so e_{1}+1=43 \Longrightarrow e_{1}=42 and e_{2}+1=47 \Longrightarrow e_{2}=46. To minimize this integer, we set p_{1}=3 and p_{2}=2. Then this integer is 3^{42} \cdot 2^{46}=2^{4} \cdot 2^{42} \cdot 3^{42}=16 \cdot 6^{42}. Now m=16 and k=42 so m+k=16+42=58=B

Solution 2:

Recall that 6^{k} can be written as 2^{k} \cdot 3^{k}. Since we want the integer to have 2021 divisors, we must have it in the form p_{1}^{42} \cdot p_{2}^{46}, where p_{1} and p_{2} are prime numbers. Therefore, we want p_{1} to be 3 and p_{2} to be 2 . To make up the remaining 2^{4}, we multiply 2^{42} \cdot 3^{42} by m, which is 2^{4} which is 16 . Therefore, we have 42+16=(B) 58

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