Nine Divisors

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Intuition

Count numbers up to n that have exactly 9 divisors. Numbers of form p^8 or p^2*q^2.

Algorithm

1Sieve primes up to sqrt(n). Count: p^8 <= n gives one form. p^2*q^2 <= n where p!=q gives another. Combine.

Common Pitfalls

•Exactly 9 divisors: n=p^8 (divisors 1,p,...,p^8) or n=p^2*q^2 (divisors: 3*3=9). Use sieve.

Nine Divisors.java

Java

// Approach: Count numbers up to n with exactly 9 divisors: form p^8 or p^2*q^2 (distinct primes).
// Time: O(sqrt(n) * log log n) Space: O(sqrt(n))
class Solution {
    public static int countNumbers(int n) {
        int ans = 0;
        for (int i = 6; i * i <= n; i++) {
            int x = i * i, cnt = 1;
            for (int j = 2; j * j < x; j++) {
                if (x % j == 0)
                    cnt++;
                if (cnt > 4)
                    break;
            }
            if (cnt == 4)
                ans++;
        }
        
        return ans;
    }
}

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