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Question

Difficulty: MediumFactors, Multiples, and Prime Factorization

What is the smallest positive integer that has exactly 2121 positive divisors?

Answer: 576

Answer

The smallest positive integer with exactly 21 positive divisors is 576.
The number 576 factors completely into 26×322^6 \times 3^2. According to the divisor formula, we take the exponents, add 1 to each, and multiply them: (6+1)×(2+1)=7×3=21(6+1) \times (2+1) = 7 \times 3 = 21. Because 2 is the smallest prime and is paired with the largest exponent, 576 is the absolute minimum positive integer that can produce exactly 21 divisors.

Step-by-Step Solution

1
Use the divisor counting formula.
The number of divisors of N=pa×qbN = p^a \times q^b is (a+1)(b+1)(a+1)(b+1). We set this equal to 2121.
This establishes the relationship between prime exponents and the given divisor count.
2
Find the factor pairs of 21.
The ways to factor 21 are 2121 (as 21×121 \times 1) or 7×37 \times 3.
This determines the possible combinations of (a+1)(a+1) and (b+1)(b+1).
3
Test the first case (single prime factor).
a+1=21  ⟹  a=20a+1 = 21 \implies a = 20. The smallest number is 220=1,048,5762^{20} = 1,048,576.
We must evaluate all cases to ensure we find the absolute minimum.
4
Test the second case (two prime factors).
(a+1)(b+1)=7×3  ⟹  a=6,b=2(a+1)(b+1) = 7 \times 3 \implies a = 6, b = 2. The number form is p6×q2p^6 \times q^2.
This evaluates the only other valid prime exponent combination.
5
Minimize the two-prime case.
Assign p=2p=2 and q=3q=3 to get 26×32=64×9=5762^6 \times 3^2 = 64 \times 9 = 576.
Assigning the smallest available prime to the largest exponent minimizes the product.
6
Compare the results of both cases.
576<1,048,576576 < 1,048,576.
To conclusively identify the smallest positive integer.

Key Concept

Prime Factorization and the Divisor Counting Formula
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Topics

Factors, Multiples, and Prime Factorization

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