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 NNN 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 NNN, and we need to find how many integers from 1 to NNN 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 xxx 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×⋯×pkekwhere 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)d(x)d(x) of xxx 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=3×39 = 3 \times 39=3×3, 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:
Perfect eighth powers of primes (like p8p^8p8).
Products of two distinct squares of primes (like p12×p22p_1^2 \times p_2^2p12×p22).
Step 2: Algorithm
For p8p^8p8 type numbers: We check for prime numbers ppp and calculate p8p^8p8. If p8p^8p8 is less than or equal to NNN, we count it.
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 NNN, 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
python
import mathdefsieve_of_eratosthenes(limit):
“”” Returns a list of primes up to limit using the Sieve of Eratosthenes. “””
sieve = [True] * (limit + 1)
sieve[0] = sieve[1] = Falsefor start inrange(2, int(math.sqrt(limit)) + 1):
if sieve[start]:
for i inrange(start*start, limit + 1, start):
sieve[i] = Falsereturn [num for num, is_prime inenumerate(sieve) if is_prime]defcount_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^8for p in primes:
if p**8 <= N:
count += 1# Case 2: Check for numbers of the form p1^2 * p2^2for i inrange(len(primes)):
for j inrange(i+1, len(primes)):
if primes[i]**2 * primes[j]**2 <= N:
count += 1return count# Example usage
N = 100print(count_numbers_with_9_divisors(N)) # Output: 2 (for 36 and 100)
Explanation of the Code
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.
Case 1: We loop through all primes and check if p8≤Np^8 \leq Np8≤N. If true, we increment the count.
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.
Efficiency: The Sieve of Eratosthenes runs in O(NloglogN)O(N \log \log N)O(NloglogN), 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
For p8p^8p8: Check for primes ppp such that p8≤100p^8 \leq 100p8≤100.
Only p=2p = 2p=2 works, because 28=2562^8 = 25628=256, which is greater than 100.
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 NNN. 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 NNN.SEO Keywords: Number of divisors, 9 divisors, Sieve of Eratosthenes, prime factorization, number of divisors algorithm.