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  3. Project Euler #5: Smallest multiple
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Project Euler #5: Smallest multiple

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  • pulla_nithish_k1
    3 weeks ago+ 0 comments

    import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

    public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for(int i = 0; i < t; i++){
            int n = in.nextInt();
            int ans = smallestDivisibleNumber(n);
            System.out.println(ans);
    
        }
    }
     private static int smallestDivisibleNumber(int n) {
        int ans = 1;
        for (int i = 2; i <= n; i++) {
            ans = lcm(ans, i);
        }
        return ans;
    }
    
    private static int lcm(int a, int b) {
        return (a * b) / gcd(a, b);
    }
    
    private static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
    

    }

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  • macdonald_l_kyn
    1 month ago+ 0 comments

    Python Solution, I cut some factor processing through prime checking

    import math
    
    
    def isPrime(n):
        if n == 1:
            return False
        if n % 2 == 0:
            return False
        for i in range(3, int(math.sqrt(n))+1, 2):
            if n % i == 0:
                return False
        return True
    
    
    for t in range(int(input())):
        n = int(input())
        s = n
        while True:
            if not isPrime(s):
                for i in range(2, n + 1):
                    if s % i != 0:
                        break
                else:
                    print(s)
                    break
            if s % 2 == 1:
                s += 1
            else:
                s += 2
    
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  • xpresskaran98
    1 month ago+ 0 comments

    python 3

    t = int(input().strip())
    for a0 in range(t):
        n = int(input().strip())
        num = 0
        while True:
            num += 1
            isOutput = True
            for i in range(1, n + 1):
                if num % i != 0:
                    isOutput = False
                    break
            if isOutput:
                break
        print(num)
    
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  • MuhammedAydogan
    2 months ago+ 0 comments

    100 points in just a millisecond:

    1) find prime numbers from 1 to N. 2) find maximum power of prime number. 3) multiply each prime powers.

    Hint: You dont need to calculate nonprime numbers because you always have enough primes to make them up

    int result = 1;
    for (int i = 1; i <= n; i++) {
      if (isPrime(i)) {
        int r1 = 1;
        while (r1 * i <= n)
          r1 *= i;
        result *= r1;
      }
    }
    return result;
    
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  • vipin_katiyar
    2 months ago+ 0 comments
    import java.math.BigInteger;
    import java.util.Scanner;
    
    public class Main {
        public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            System.out.println(lcm(n));
        }
        sc.close();
    }
    
    private static BigInteger lcm(int n) {
        BigInteger lcm = BigInteger.valueOf(1);
        for (int i = 2; i <= n; i++) {
            lcm = lcm(lcm, i);
        }
        return lcm;
    }
    
    private static BigInteger lcm(BigInteger a, int b) {
        return a.multiply(BigInteger.valueOf(b))
                 .divide(a.gcd(BigInteger.valueOf(b)));
    }
    }
    
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