Question

Difficulty: HardPrime Numbers and Prime Factorization

A positive integer nn has exactly three distinct prime factors. If n2n^2 has exactly 147147 positive divisors, what is the minimum possible value of nn?

Answer: 1080

Answer

The minimum possible value of nn is 1080.
By prime factorizing n=paqbrcn = p^a q^b r^c, the number of divisors of n2n^2 is given by (2a+1)(2b+1)(2c+1)=147(2a+1)(2b+1)(2c+1) = 147. Decomposing 147 into three factors greater than 1 gives 7×7×37 \times 7 \times 3, which determines the exponents a=3a=3, b=3b=3, and c=1c=1. Minimizing nn by assigning the smallest primes (2, 3, and 5) to these exponents yields 23×33×51=10802^3 \times 3^3 \times 5^1 = 1080.

Step-by-Step Solution

1
Set up the prime factorization of nn and n2n^2
Let n=paqbrcn = p^a q^b r^c, where p<q<rp < q < r are prime numbers and a,b,c1a, b, c \ge 1. Then n2=p2aq2br2cn^2 = p^{2a} q^{2b} r^{2c}.
Any positive integer can be uniquely represented as a product of prime powers.
2
Formulate the divisor counting equation for n2n^2
The total number of positive divisors of n2n^2 is (2a+1)(2b+1)(2c+1)=147(2a+1)(2b+1)(2c+1) = 147.
If an integer has prime factorization piei\prod p_i^{e_i}, its number of positive divisors is (ei+1)\prod (e_i + 1).
3
Factor 147 into three odd factors greater than 1
The prime factorization of 147 is 3×723 \times 7^2. The only factorization into three integers greater than 1 is 7×7×37 \times 7 \times 3.
Since a,b,c1a, b, c \ge 1, we have 2a+132a+1 \ge 3, 2b+132b+1 \ge 3, and 2c+132c+1 \ge 3.
4
Solve for the exponents a,b,ca, b, c
Setting 2a+1=72a+1=7, 2b+1=72b+1=7, and 2c+1=32c+1=3 yields a=3a=3, b=3b=3, and c=1c=1.
Equating the factor values allows us to determine the required exponents of the prime factors of nn.
5
Minimize nn by choosing prime bases and assigning exponents
Choose the three smallest primes p=2,q=3,r=5p=2, q=3, r=5 and assign exponents 3,3,13, 3, 1 to get n=23×33×51=1080n = 2^3 \times 3^3 \times 5^1 = 1080.
To make the product as small as possible, smaller prime bases must be paired with larger exponents.

Key Concept

Determining the prime factorization and number of divisors of prime power expressions
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