A number N is said to be Primitive Abundant Number if N is an Abundant number and all it’s proper divisors are Deficient Numbers.
The first few Primitive Abundant Numbers are:
20, 70, 88, 104, 272, 304………
Check if N is a Primitive Abundant Number
Given a number N, the task is to find if this number is Primitive Abundant Number or not.
Examples:
Input: N = 20
Output: YES
Explanation:
Sum of 20’s proper divisors is – 1 + 2 + 4 + 5 + 10 = 22 > 20,
So, 20 is an abundant number.
The proper divisors of 1, 2, 4, 5 and 10 are0, 1, 3, 1 and 8 respectively,
Each of these numbers is a deficient number
Therefore, 20 is a primitive abundant number.
Input: N = 17
Output: No
Approach:
- Check if the number is an Abundant number or not, i.e, sum of all the proper divisors of the number denoted by sum(N) is greater than the value of the number N
- If the number is not abundant then return false else do the following
- Check if all proper divisors of N are Deficient Numbers or not, i.e, sum of all the divisors of the number denoted by divisorsSum(n) is less than twice the value of the number N.
- If both above conditions are true print “Yes” else print “No.
Below is the implementation of the above approach:
C++
// C++ implementation of the above// approach#include <bits/stdc++.h>using namespace std;// Function to sum of divisorsint getSum(int n){ int sum = 0; // Note that this loop // runs till square root of N for (int i = 1; i <= sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // take only one of them if (n / i == i) sum = sum + i; else // Otherwise take both { sum = sum + i; sum = sum + (n / i); } } } return sum;}// Function to check Abundant Numberbool checkAbundant(int n){ // Return true if sum // of divisors is greater // than N. return (getSum(n) - n > n);}// Function to check Deficient Numberbool isDeficient(int n){ // Check if sum(n) < 2 * n return (getSum(n) < (2 * n));}// Function to check all proper divisors// of N is deficient number or notbool checkPrimitiveAbundant(int num){ // if number itself is not abundant // return false if (!checkAbundant(num)) { return false; } // find all divisors which divides 'num' for (int i = 2; i <= sqrt(num); i++) { // if 'i' is divisor of 'num' if (num % i == 0 && i != num) { // if both divisors are same then add // it only once else add both if (i * i == num) { if (!isDeficient(i)) { return false; } } else if (!isDeficient(i) || !isDeficient(num / i)) { return false; } } } return true;}// Driver Codeint main(){ int n = 20; if (checkPrimitiveAbundant(n)) { cout << "Yes"; } else { cout << "No"; } return 0;} |
Java
// Java implementation of the above// approachclass GFG{ // Function to sum of divisorsstatic int getSum(int n){ int sum = 0; // Note that this loop runs // till square root of N for(int i = 1; i <= Math.sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // take only one of them if (n / i == i) sum = sum + i; // Otherwise take both else { sum = sum + i; sum = sum + (n / i); } } } return sum;}// Function to check Abundant Numberstatic boolean checkAbundant(int n){ // Return true if sum // of divisors is greater // than N. return (getSum(n) - n > n);}// Function to check Deficient Numberstatic boolean isDeficient(int n){ // Check if sum(n) < 2 * n return (getSum(n) < (2 * n));}// Function to check all proper divisors// of N is deficient number or notstatic boolean checkPrimitiveAbundant(int num){ // If number itself is not abundant // return false if (!checkAbundant(num)) { return false; } // Find all divisors which divides 'num' for(int i = 2; i <= Math.sqrt(num); i++) { // if 'i' is divisor of 'num' if (num % i == 0 && i != num) { // if both divisors are same then // add it only once else add both if (i * i == num) { if (!isDeficient(i)) { return false; } } else if (!isDeficient(i) || !isDeficient(num / i)) { return false; } } } return true;}// Driver Codepublic static void main(String[] args) { int n = 20; if (checkPrimitiveAbundant(n)) { System.out.print("Yes"); } else { System.out.print("No"); }}}// This code is contributed by Ritik Bansal |
Python3
# Python3 implementation of the above# approachimport math# Function to sum of divisorsdef getSum(n): sum = 0 # Note that this loop # runs till square root of N for i in range(1, int(math.sqrt(n) + 1)): if (n % i == 0): # If divisors are equal, # take only one of them if (n // i == i): sum = sum + i else: # Otherwise take both sum = sum + i sum = sum + (n // i) return sum# Function to check Abundant Numberdef checkAbundant(n): # Return True if sum # of divisors is greater # than N. if (getSum(n) - n > n): return True return False# Function to check Deficient Numberdef isDeficient(n): # Check if sum(n) < 2 * n if (getSum(n) < (2 * n)): return True return False# Function to check all proper divisors# of N is deficient number or notdef checkPrimitiveAbundant(num): # if number itself is not abundant # return False if not checkAbundant(num): return False # find all divisors which divides 'num' for i in range(2, int(math.sqrt(num) + 1)): # if 'i' is divisor of 'num' if (num % i == 0 and i != num): # if both divisors are same then add # it only once else add both if (i * i == num): if (not isDeficient(i)): return False elif (not isDeficient(i) or not isDeficient(num // i)): return False return True# Driver Coden = 20if (checkPrimitiveAbundant(n)): print("Yes")else: print("No")# This code is contributed by shubhamsingh10 |
C#
// C# implementation of the above// approachusing System;class GFG{ // Function to sum of divisorsstatic int getSum(int n){ int sum = 0; // Note that this loop runs // till square root of N for(int i = 1; i <= Math.Sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // take only one of them if (n / i == i) sum = sum + i; // Otherwise take both else { sum = sum + i; sum = sum + (n / i); } } } return sum;}// Function to check Abundant Numberstatic bool checkAbundant(int n){ // Return true if sum // of divisors is greater // than N. return (getSum(n) - n > n);}// Function to check Deficient Numberstatic bool isDeficient(int n){ // Check if sum(n) < 2 * n return (getSum(n) < (2 * n));}// Function to check all proper divisors// of N is deficient number or notstatic bool checkPrimitiveAbundant(int num){ // If number itself is not abundant // return false if (!checkAbundant(num)) { return false; } // Find all divisors which divides 'num' for(int i = 2; i <= Math.Sqrt(num); i++) { // If 'i' is divisor of 'num' if (num % i == 0 && i != num) { // If both divisors are same then // add it only once else add both if (i * i == num) { if (!isDeficient(i)) { return false; } } else if (!isDeficient(i) || !isDeficient(num / i)) { return false; } } } return true;}// Driver Codepublic static void Main() { int n = 20; if (checkPrimitiveAbundant(n)) { Console.Write("Yes"); } else { Console.Write("No"); }}}// This code is contributed by Code_Mech |
Javascript
<script>// Javascript implementation of the above// approach // Function to sum of divisors function getSum( n) { let sum = 0; // Note that this loop runs // till square root of N for ( let i = 1; i <= Math.sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // take only one of them if (n / i == i) sum = sum + i; // Otherwise take both else { sum = sum + i; sum = sum + (n / i); } } } return sum; } // Function to check Abundant Number function checkAbundant( n) { // Return true if sum // of divisors is greater // than N. return (getSum(n) - n > n); } // Function to check Deficient Number function isDeficient( n) { // Check if sum(n) < 2 * n return (getSum(n) < (2 * n)); } // Function to check all proper divisors // of N is deficient number or not function checkPrimitiveAbundant( num) { // If number itself is not abundant // return false if (!checkAbundant(num)) { return false; } // Find all divisors which divides 'num' for ( let i = 2; i <= Math.sqrt(num); i++) { // if 'i' is divisor of 'num' if (num % i == 0 && i != num) { // if both divisors are same then // add it only once else add both if (i * i == num) { if (!isDeficient(i)) { return false; } } else if (!isDeficient(i) || !isDeficient(num / i)) { return false; } } } return true; } // Driver Code let n = 20; if (checkPrimitiveAbundant(n)) { document.write("Yes"); } else { document.write("No"); }// This code contributed by aashish1995 </script> |
Output
Yes
Time Complexity: O(N1/2)
Auxiliary Space: O(1)
Approach 2: Dynamic Programming:
- In this approach, we use dynamic programming to memoize the results of whether a number is abundant or not. We first initialize a DP array dp with all elements set to -1 to indicate that we haven’t computed the result for that number yet.
- We then define a sum_of_divisors function that computes the sum of divisors of a given number. We use this function to check whether a number is abundant or not. If the result for a particular number n is already present in the DP array, we return that result directly. Otherwise, we compute the result using the sum_of_divisors function and store it in the DP array for future use.
- Finally, we check if the given number is abundant or not using the is_abundant function and print “Yes” or “No” accordingly.
Here is the code below:
C++
#include<bits/stdc++.h>using namespace std;const int N = 1e5 + 5;int dp[N];// Function to calculate sum of divisorsint sum_of_divisors(int n) { int sum = 1; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { sum += i; if (n / i != i) { sum += n / i; } } } return sum;}// Function to check if a number is Abundant or notbool checkPrimitiveAbundant(int n) { if (dp[n] != -1) { return dp[n]; } return dp[n] = (sum_of_divisors(n) > n);}int main() { memset(dp, -1, sizeof(dp)); int n = 20; if (checkPrimitiveAbundant(n)) { cout << "Yes"; } else { cout << "No"; } return 0;} |
Java
import java.util.*;public class Main { static final int N = 100000; static int[] dp = new int[N]; // Function to calculate sum of divisors static int sumOfDivisors(int n) { int sum = 1; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { sum += i; if (n / i != i) { sum += n / i; } } } return sum; } // Function to check if a number is Abundant or not static boolean checkPrimitiveAbundant(int n) { if (dp[n] != 0) { return dp[n] == 1; } return (dp[n] = (sumOfDivisors(n) > n) ? 1 : -1) == 1; } public static void main(String[] args) { Arrays.fill(dp, 0); int n = 20; if (checkPrimitiveAbundant(n)) { System.out.println("Yes"); } else { System.out.println("No"); } }} |
Python
N = 100005dp = [-1] * N# Function to calculate sum of divisorsdef sum_of_divisors(n): sum = 1 for i in range(2, int(n**0.5) + 1): if n % i == 0: sum += i if n // i != i: sum += n // i return sum# Function to check if a number is Abundant or notdef checkPrimitiveAbundant(n): if dp[n] != -1: # Check if the result is already calculated and stored in dp array return dp[n] dp[n] = sum_of_divisors(n) > n # Store the result of abundant check in dp array return dp[n]if __name__ == "__main__": n = 20 if checkPrimitiveAbundant(n): print("Yes") # If n is primitive abundant, print "Yes" else: print("No") # Otherwise, print "No" |
C#
using System;public class GFG{ const int N = 100000; static int[] dp = new int[N]; // Function to calculate sum of divisors static int SumOfDivisors(int n) { int sum = 1; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { sum += i; if (n / i != i) { sum += n / i; } } } return sum; } // Function to check if a number is Abundant or not static bool CheckPrimitiveAbundant(int n) { if (dp[n] != 0) { return dp[n] == 1; } return (dp[n] = (SumOfDivisors(n) > n) ? 1 : -1) == 1; } public static void Main(string[] args) { Array.Fill(dp, 0); int n = 20; if (CheckPrimitiveAbundant(n)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This Code is Contributed by Dwaipayan Bandyopadhyay |
Javascript
const N = 1e5 + 5;let dp = new Array(N);// Function to calculate sum of divisorsfunction sum_of_divisors(n) { let sum = 1; for (let i = 2; i * i <= n; i++) { if (n % i == 0) { sum += i; if (n / i != i) { sum += n / i; } } } return sum;}// Function to check if a number is Abundant or notfunction checkPrimitiveAbundant(n) { if (dp[n] != undefined) { return dp[n]; } return dp[n] = (sum_of_divisors(n) > n);}let n = 20;if (checkPrimitiveAbundant(n)) { console.log("Yes");} else { console.log("No");} |
Output:
Yes
Time Complexity: O(N log log N), where N is the input number.
Auxiliary Space: O(N), as we need to create an array of size N to store the sum of divisors for all numbers from 1 to N.
References: https://en.wikipedia.org/wiki/Primitive_abundant_number
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