Question

Difficulty: HardDivisibility, Factors, and Multiples

A positive integer nn is not divisible by 33. If nn has exactly 1515 positive divisors and 2n2n has exactly 2020 positive divisors, how many positive divisors does 3n3n have?

Answer: 30

Answer

30
The number of positive divisors of an integer with prime factorization p1e1p2e2…p_1^{e_1} p_2^{e_2} \dots is (e1+1)(e2+1)…(e_1 + 1)(e_2 + 1) \dots. Given d(n)=15d(n) = 15, nn can be p14p^{14} or p4q2p^4 q^2. Given d(2n)=20d(2n) = 20, 2 must be a prime factor of nn with exponent 2, making n=22p4n = 2^2 p^4. Since nn is not divisible by 3, p≠3p \neq 3. Multiplying nn by 3 introduces 313^1 into the prime factorization, giving 3n=22⋅31⋅p43n = 2^2 \cdot 3^1 \cdot p^4. The total number of positive divisors is (2+1)(1+1)(4+1)=30(2+1)(1+1)(4+1) = 30.

Step-by-Step Solution

1
Analyze the number of positive divisors of nn.
nn is either p14p^{14} or p4q2p^4 q^2 for distinct primes pp and qq.
The number of positive divisors of n=p1e1p2e2…n = p_1^{e_1} p_2^{e_2} \dots is given by (e1+1)(e2+1)…(e_1 + 1)(e_2 + 1) \dots. Since 15=15×1=5×315 = 15 \times 1 = 5 \times 3, the exponent structure must be 14 or 4 and 2.
2
Analyze the number of positive divisors of 2n2n.
n=22p4n = 2^2 p^4, where pp is a prime other than 2 and 3.
If n=p4q2n = p^4 q^2, multiplying by 2 increases the exponent of 2 by 1. If q=2q = 2, then n=22p4n = 2^2 p^4 and 2n=23p42n = 2^3 p^4, yielding (3+1)(4+1)=20(3 + 1)(4 + 1) = 20 divisors. Any other prime structure yields a different count.
3
Determine the prime factorization of 3n3n and count its positive divisors.
The number of divisors of 3n3n is 30.
Since nn is not divisible by 3, p≠3p \neq 3. Thus 3n=22⋅31⋅p43n = 2^2 \cdot 3^1 \cdot p^4, which has (2+1)(1+1)(4+1)=30(2 + 1)(1 + 1)(4 + 1) = 30 positive divisors.

Key Concept

Divisor Count Formula and Prime Factorization
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