Find Numbers with Exactly 9 Divisors in a Range | Efficient Approach

In this problem, we need to identify how many integers within a given range $N$ have exactly 9 divisors. Let’s break down the problem and explore an efficient approach to solve it.

Problem Overview

We are given an integer $N$, and we need to find how many integers from 1 to $N$ have exactly 9 divisors. For example, the number 36 has 9 divisors: 1, 2, 3, 4, 6, 9, 12, 18, and 36. Similarly, 100 also has 9 divisors: 1, 2, 4, 5, 10, 20, 25, 50, and 100.

Approach

Step 1: Understanding Divisors

The number of divisors of a number can be determined by its prime factorization. If a number $x$ has the prime factorization:

x=p1e1×p2e2×⋯×pkekx = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}x=p1e1​​×p2e2​​×⋯×pkek​​

where p1,p2,…,pkp_1, p_2, \dots, p_kp1​,p2​,…,pk​ are distinct prime numbers and e1,e2,…,eke_1, e_2, \dots, e_ke1​,e2​,…,ek​ are the respective exponents, the number of divisors $d(x)$ of $x$ is given by:

d(x)=(e1+1)×(e2+1)×⋯×(ek+1)d(x) = (e_1 + 1) \times (e_2 + 1) \times \cdots \times (e_k + 1)d(x)=(e1​+1)×(e2​+1)×⋯×(ek​+1)

For a number to have exactly 9 divisors, we need the product of these terms to be equal to 9. The factorizations of 9 are:

• $9=9\times 1$, which implies e1=8e_1 = 8e1​=8 (i.e., x=p18x = p_1^8x=p18​).
• $9=3\times 3$, which implies e1=2e_1 = 2e1​=2 and e2=2e_2 = 2e2​=2 (i.e., x=p12×p22x = p_1^2 \times p_2^2x=p12​×p22​).

Thus, the two types of numbers with exactly 9 divisors are:

1. Perfect eighth powers of primes (like p8p^8p8).
2. Products of two distinct squares of primes (like p12×p22p_1^2 \times p_2^2p12​×p22​).

Step 2: Algorithm

1. For p8p^8p8 type numbers: We check for prime numbers $p$ and calculate p8p^8p8. If p8p^8p8 is less than or equal to $N$, we count it.
2. For p12×p22p_1^2 \times p_2^2p12​×p22​ type numbers: We check all pairs of prime numbers p1p_1p1​ and p2p_2p2​ and calculate p12×p22p_1^2 \times p_2^2p12​×p22​. If the result is less than or equal to $N$, we count it.

Step 3: Efficient Calculation

To efficiently find primes, we can use the Sieve of Eratosthenes to generate all primes up to N\sqrt{N}N​. This will help us generate the required primes for both types of numbers.

Python Code Implementation

Explanation of the Code

1. Sieve of Eratosthenes: The sieve_of_eratosthenes function generates all prime numbers up to N\sqrt{N}N​. This is essential for finding the prime factors needed for both types of numbers with 9 divisors.
2. Case 1: We loop through all primes and check if p8≤Np^8 \leq Np8≤N. If true, we increment the count.
3. Case 2: We loop through all pairs of primes p1p_1p1​ and p2p_2p2​, and check if p12×p22≤Np_1^2 \times p_2^2 \leq Np12​×p22​≤N. If true, we increment the count.
4. Efficiency: The Sieve of Eratosthenes runs in $O(N\log \log N)$, and the double loop for pairs of primes is manageable due to the limited number of primes less than N\sqrt{N}N​.

Example Walkthrough

Example 1:

Input:
N = 100

1. For p8p^8p8: Check for primes $p$ such that p8≤100p^8 \leq 100p8≤100.
   • Only $p=2$ works, because 28=2562^8 = 25628=256, which is greater than 100.
2. For p12×p22p_1^2 \times p_2^2p12​×p22​: Check for pairs of primes p1p_1p1​ and p2p_2p2​.
   • 22×32=362^2 \times 3^2 = 3622×32=36, which is valid.
   • 22×52=1002^2 \times 5^2 = 10022×52=100, which is valid.

Thus, the numbers with exactly 9 divisors are 36 and 100.

Output:
2

Conclusion

This approach efficiently solves the problem of finding how many integers have exactly 9 divisors in a given range $N$. The solution leverages the Sieve of Eratosthenes to find primes and then checks the two possible forms of numbers with 9 divisors: p8p^8p8 and p12×p22p_1^2 \times p_2^2p12​×p22​. This solution works within the problem’s constraints and is optimized for large values of $N$.