Primitive Abundant Number (original) (raw)
Last Updated : 15 Jul, 2025
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 divisors int 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 Number bool checkAbundant(int n) { // Return true if sum // of divisors is greater // than N. return (getSum(n) - n > n); }
// Function to check Deficient Number 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 not 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 <= 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 int main() {
int n = 20;
if (checkPrimitiveAbundant(n)) {
cout << "Yes";
}
else {
cout << "No";
}
return 0;}
Java
// Java implementation of the above // approach class GFG{
// Function to sum of divisors static 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 Number static boolean checkAbundant(int n) {
// Return true if sum
// of divisors is greater
// than N.
return (getSum(n) - n > n);}
// Function to check Deficient Number static 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 not static 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 Code public 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
approach
import math
Function to sum of divisors
def 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 sumFunction to check Abundant Number
def checkAbundant(n):
# Return True if sum
# of divisors is greater
# than N.
if (getSum(n) - n > n):
return True
return FalseFunction to check Deficient Number
def isDeficient(n):
# Check if sum(n) < 2 * n
if (getSum(n) < (2 * n)):
return True
return FalseFunction to check all proper divisors
of N is deficient number or not
def 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 TrueDriver Code
n = 20 if (checkPrimitiveAbundant(n)): print("Yes") else: print("No")
This code is contributed by shubhamsingh10
C#
// C# implementation of the above // approach using System; class GFG{
// Function to sum of divisors static 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 Number static bool checkAbundant(int n) {
// Return true if sum
// of divisors is greater
// than N.
return (getSum(n) - n > n);}
// Function to check Deficient Number static 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 not static 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 Code public static void Main() { int n = 20;
if (checkPrimitiveAbundant(n))
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}} }
// This code is contributed by Code_Mech
JavaScript
`
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 divisors int 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 not bool 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 = 100005 dp = [-1] * N
Function to calculate sum of divisors
def 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 not
def 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 divisors function 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 not function 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