Finding the Number of Integers with Exactly 9 Divisors in a Given Range

Finding the Number of Integers with Exactly 9 Divisors in a Given Range

In this problem, we need to identify how many integers within a given range NN have exactly 9 divisors. Let’s break down the problem and explore an efficient approach to solve it.Finding the Number of Integers with Exactly 9 Divisors in a Given Range

Problem Overview

We are given an integer NN, and we need to find how many integers from 1 to NN 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 xx has the prime factorization:x=p1e1×p2e2×⋯×pkekx = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}where p1,p2,…,pkp_1, p_2, \dots, p_k are distinct prime numbers and e1,e2,…,eke_1, e_2, \dots, e_k are the respective exponents, the number of divisors d(x)d(x) of xx 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)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=9×19 = 9 \times 1, which implies e1=8e_1 = 8 (i.e., x=p18x = p_1^8).
  • 9=3×39 = 3 \times 3, which implies e1=2e_1 = 2 and e2=2e_2 = 2 (i.e., x=p12×p22x = p_1^2 \times p_2^2).
Thus, the two types of numbers with exactly 9 divisors are:
  1. Perfect eighth powers of primes (like p8p^8).
  2. Products of two distinct squares of primes (like p12×p22p_1^2 \times p_2^2).

Step 2: Algorithm

  1. For p8p^8 type numbers: We check for prime numbers pp and calculate p8p^8. If p8p^8 is less than or equal to NN, we count it.
  2. For p12×p22p_1^2 \times p_2^2 type numbers: We check all pairs of prime numbers p1p_1 and p2p_2 and calculate p12×p22p_1^2 \times p_2^2. If the result is less than or equal to NN, 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}. This will help us generate the required primes for both types of numbers.

Python Code Implementation

python
import mathdef sieve_of_eratosthenes(limit): “”” Returns a list of primes up to limit using the Sieve of Eratosthenes. “”” sieve = [True] * (limit + 1) sieve[0] = sieve[1] = False for start in range(2, int(math.sqrt(limit)) + 1): if sieve[start]: for i in range(start*start, limit + 1, start): sieve[i] = False return [num for num, is_prime in enumerate(sieve) if is_prime]def count_numbers_with_9_divisors(N): count = 0 limit = int(math.sqrt(N)) # Sieve limit for prime factors# Get all primes up to sqrt(N) primes = sieve_of_eratosthenes(limit)# Case 1: Check for numbers of the form p^8 for p in primes: if p**8 <= N: count += 1# Case 2: Check for numbers of the form p1^2 * p2^2 for i in range(len(primes)): for j in range(i+1, len(primes)): if primes[i]**2 * primes[j]**2 <= N: count += 1return count# Example usage N = 100 print(count_numbers_with_9_divisors(N)) # Output: 2 (for 36 and 100)

Explanation of the Code

  1. Sieve of Eratosthenes: The sieve_of_eratosthenes function generates all prime numbers up to N\sqrt{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 N. If true, we increment the count.
  3. Case 2: We loop through all pairs of primes p1p_1 and p2p_2, and check if p12×p22≤Np_1^2 \times p_2^2 \leq N. If true, we increment the count.
  4. Efficiency: The Sieve of Eratosthenes runs in O(Nlog⁡log⁡N)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}.

Example Walkthrough

Example 1:

Input: N = 100
  1. For p8p^8: Check for primes pp such that p8≤100p^8 \leq 100.
    • Only p=2p = 2 works, because 28=2562^8 = 256, which is greater than 100.
  2. For p12×p22p_1^2 \times p_2^2: Check for pairs of primes p1p_1 and p2p_2.
    • 22×32=362^2 \times 3^2 = 36, which is valid.
    • 22×52=1002^2 \times 5^2 = 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 NN. The solution leverages the Sieve of Eratosthenes to find primes and then checks the two possible forms of numbers with 9 divisors: p8p^8 and p12×p22p_1^2 \times p_2^2. This solution works within the problem’s constraints and is optimized for large values of NN.SEO Keywords: Number of divisors, 9 divisors, Sieve of Eratosthenes, prime factorization, number of divisors algorithm.
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