A number n is said to be an Abundant Number if the sum of all the proper divisors of the number denoted by sum(n) is greater than the value of the number n. And the difference between these two values is called abundance.
Mathematically, if the below condition holds the number is said to be an Abundant number:
sum(n)> n
abundance = sum(n) - n
sum(n): aliquot sum - The sum of all proper divisors of n
Given a number n, our task is to find if this number is an Abundant number or not.
The first few Abundant Numbers are: 12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66 .....
Examples:
Input: 21
Output: NO
Input: 12
Output: YES
Input: 17
Output: NO
Method 1: A Simple solution is to iterate all the numbers from 1 to n-1 and check if the number divides n and calculate the sum. Check if this sum is greater than n or not.
C++
// C++ program to find if a given
// number is Abundant number or not.
#include <bits/stdc++.h>
using namespace std;
// Returns true if the given number is Abundant
bool isAbundantNumber(int n)
{
// To store the sum of divisors
int sum = 0;
// Loop through the numbers [1,n-1]
for (int i = 1; i < n; i++) {
if (n % i == 0) {
sum += i;
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
if (sum > n) {
return true;
}
else {
return false;
}
}
// Driver program
int main()
{
// Function call
if (isAbundantNumber(12)) {
cout << "YES" << endl;
;
}
else {
cout << "NO" << endl;
;
}
}
// This code is contributed by phasing17
// Java program to find if a given
// number is Abundant number or not.
import java.io.*;
import java.util.*;
class GFG {
// Returns true if the given number is Abundant
public static boolean isAbundantNumber(int n)
{
// To store the sum of divisors
int sum = 0;
// Loop through the numbers [1,n-1]
for (int i = 1; i < n; i++) {
if (n % i == 0) {
sum += i;
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
if (sum > n) {
return true;
}
else {
return false;
}
}
// Driver program
public static void main(String[] args)
{
// Function call
if (isAbundantNumber(12)) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}
// This code is contributed by shruti456rawal
# code
n = 12
s = 0
# iterating loop n times
for i in range(1, n):
# finding proper divisors
if n % i == 0:
# adding proper divisors to the sum s
s += i
# checking if sum is greater than the
# given number then it is called
# anundant so print yes otherwise no
if s > n:
print("Yes")
else:
print("No")
# this code is contributed by gangarajula laxmi
// C# program to find if a given
// number is Abundant number or not.
using System;
class GFG {
// Returns true if the given number is Abundant
public static bool isAbundantNumber(int n)
{
// To store the sum of divisors
int sum = 0;
// Loop through the numbers [1,n-1]
for (int i = 1; i < n; i++) {
if (n % i == 0) {
sum += i;
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
if (sum > n) {
return true;
}
else {
return false;
}
}
// Driver program
public static void Main(string[] args)
{
// Function call
if (isAbundantNumber(12)) {
Console.WriteLine("YES");
}
else {
Console.WriteLine("NO");
}
}
}
// This code is contributed by phasing17
<script>
// JavaScript code for the above approach
// Returns true if the given number is Abundant
function isAbundantNumber(n) {
// To store the sum of divisors
let sum = 0;
// Loop through the numbers [1,n-1]
for (let i = 1; i < n; i++) {
if (n % i == 0) {
sum += i;
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
if (sum > n) {
return true;
}
else {
return false;
}
}
// Driver program
// Function call
if (isAbundantNumber(12)) {
document.write("YES");
}
else {
document.write("NO");
}
// This code is contributed by Potta Lokesh
</script>
<?php
$n=12;
$s=0;
// Loop through the numbers [1,n-1]
for ($i = 1; $i < $n; $i++) {
if ($n % $i == 0) {
$s += $i;
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
if ($s > $n) {
echo "Yes" ;
}
else {
echo "No";
}
// This code is contributed by laxmigangarajula03
?>
Time Complexity: O(n) for a given number n.
Auxiliary Space: O(1)
Optimized Solution:
If we observe carefully, the divisors of the number n are present in pairs. For example if n = 100, then all the pairs of divisors are: (1,100), (2,50), (4,25), (5,20), (10,10)
Using this fact we can speed up our program. While checking divisors we will have to be careful if there are two equal divisors as in the case of (10, 10). In such a case, we will take only one of them in the calculation of the sum.
Subtract the number n from the sum of all divisors to get the sum of proper divisors.
C++
// An Optimized Solution to check Abundant Number
#include <bits/stdc++.h>
using namespace std;
// Function to calculate 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);
}
}
}
// calculate sum of all proper divisors only
sum = sum - n;
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);
}
/* Driver program to test above function */
int main()
{
checkAbundant(12)? cout << "YES\n" : cout << "NO\n";
checkAbundant(15)? cout << "YES\n" : cout << "NO\n";
return 0;
}
// An Optimized Solution to check Abundant Number
// in JAVA
import java.io.*;
import java.math.*;
// Function to calculate sum of divisors
class GFG{
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;
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all proper divisors
// only
sum = sum - n;
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);
}
/* Driver program to test above function */
public static void main(String args[])throws
IOException
{
if(checkAbundant(12))
System.out.println("YES");
else
System.out.println("NO");
if(checkAbundant(15))
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by Nikita Tiwari.
# An Optimized Solution to check Abundant Number
# in PYTHON
import math
# Function to calculate sum of divisors
def getSum(n) :
sum = 0
# Note that this loop runs till square root
# of n
i = 1
while i <= (math.sqrt(n)) :
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 )
i = i + 1
# calculate sum of all proper divisors only
sum = sum - n
return sum
# Function to check Abundant Number
def checkAbundant(n) :
# Return true if sum of divisors is greater
# than n.
if (getSum(n) > n) :
return 1
else :
return 0
# Driver program to test above function */
if(checkAbundant(12) == 1) :
print ("YES")
else :
print ("NO")
if(checkAbundant(15) == 1) :
print ("YES")
else :
print ("NO")
# This code is contributed by Nikita Tiwari.
// An Optimized Solution to check Abundant Number
// in C#
// Function to calculate sum of divisors
using System;
class GFG {
// Function to calculate 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;
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all proper divisors
// only
sum = sum - n;
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);
}
/* Driver program to test above function */
public static void Main()
{
if (checkAbundant(12))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
if (checkAbundant(15))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed by vt_m.
<script>
// Javascript program for an Optimized
// solution to check Abundant Number
// Function to calculate
// 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);
}
}
}
// calculate sum of all
// proper divisors only
sum = sum - n;
return sum;
}
// Function to check Abundant Number
function checkAbundant(n)
{
// Return true if sum of
// divisors is greater than n.
return (getSum(n) > n);
}
// Driver Code
let k = checkAbundant(12) ? "YES<br>" : "NO<br>";
document.write(k);
k = checkAbundant(15) ? "YES<br>" : "NO<br>";
document.write(k);
// This code is contributed by _saurabh_jaiswal
</script>
<?php
// PHP program for an Optimized
// solution to check Abundant Number
// Function to calculate
// sum of divisors
function getSum($n)
{
$sum = 0;
// Note that this loop runs
// till square root of n
for ($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;
// Otherwise take both
else
{
$sum = $sum + $i;
$sum = $sum + ($n / $i);
}
}
}
// calculate sum of all
// proper divisors only
$sum = $sum - $n;
return $sum;
}
// Function to check Abundant Number
function checkAbundant($n)
{
// Return true if sum of
// divisors is greater than n.
return (getSum($n) > $n);
}
// Driver Code
$k = checkAbundant(12) ? "YES\n" : "NO\n";
echo($k);
$k = checkAbundant(15) ? "YES\n" : "NO\n";
echo($k);
// This code is contributed by Ajit.
?>
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
Approach 3: Dynamic Programming:
The approach uses dynamic programming to determine if a number is abundant or not.
- The idea is to create a boolean array of size n+1, where n is the number whose abundance needs to be checked. This boolean array is used to store the result of whether a number is abundant or not. Initially, all the elements in the array are set to false.
- Then, the program loops through all the numbers from 1 to n, and for each number, it calculates the sum of its divisors. If the sum of divisors is greater than the number itself, the number is marked as abundant in the boolean array.
- Finally, the function returns the value at index n in the boolean array, which tells whether the number n is abundant or not.
- This approach uses dynamic programming because it stores the results of previously computed subproblems in the boolean array and uses them to compute the abundance of larger numbers. This avoids redundant computations and makes the program more efficient.
Here is the Code of above approach:
C++
#include <iostream>
#include <vector>
using namespace std;
// Returns true if the given number is Abundant
bool isAbundantNumber(int n)
{
// To store the sum of divisors
int sum = 1;
// Loop through the numbers [2,sqrt(n)]
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
if (i * i != n) {
sum += i + n / i;
} else {
sum += i;
}
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
if (sum > n) {
return true;
}
else {
return false;
}
}
// Function to find all abundant numbers up to n using DP
vector<int> getAllAbundantNumbers(int n)
{
vector<int> abundantNumbers;
vector<int> dp(n + 1, 0);
// Calculate the sum of divisors for all numbers [1, n]
for (int i = 1; i <= n; i++) {
for (int j = i * 2; j <= n; j += i) {
dp[j] += i;
}
}
// Find all abundant numbers up to n
for (int i = 12; i <= n; i++) {
if (dp[i] > i) {
abundantNumbers.push_back(i);
}
}
return abundantNumbers;
}
// Driver program
int main()
{
// Check if 12 is an abundant number
if (isAbundantNumber(12)) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
import java.util.ArrayList;
import java.util.List;
class GFG {
// Returns true if the given number is Abundant
static boolean isAbundantNumber(int n) {
// To store the sum of divisors
int sum = 1;
// Loop through the numbers [2, sqrt(n)]
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
if (i * i != n) {
sum += i + n / i;
} else {
sum += i;
}
}
}
// A number n is said to be Abundant Number if
// the sum of all the proper divisors of the number
// is greater than the value of the number n.
return sum > n;
}
// Function to find all abundant numbers up to n using DP
static List<Integer> getAllAbundantNumbers(int n) {
List<Integer> abundantNumbers = new ArrayList<>();
int[] dp = new int[n + 1];
// Calculate the sum of divisors for all numbers [1, n]
for (int i = 1; i <= n; i++) {
for (int j = i * 2; j <= n; j += i) {
dp[j] += i;
}
}
// Find all abundant numbers up to n
for (int i = 12; i <= n; i++) {
if (dp[i] > i) {
abundantNumbers.add(i);
}
}
return abundantNumbers;
}
// Driver program
public static void main(String[] args) {
// Check if 12 is an abundant number
if (isAbundantNumber(12)) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
// by phasing17
# Function to check if the given number is Abundant
def isAbundantNumber(n):
# To store the sum of divisors
sum_divisors = 1
# Loop through the numbers [2, sqrt(n)]
i = 2
while i * i <= n:
if n % i == 0:
if i * i != n:
sum_divisors += i + n // i
else:
sum_divisors += i
i += 1
# A number n is said to be Abundant Number if
# the sum of all proper divisors of the number
# is greater than the value of the number n.
if sum_divisors > n:
return True
else:
return False
# Function to find all abundant numbers up to n using DP
def getAllAbundantNumbers(n):
abundantNumbers = []
dp = [0] * (n + 1)
# Calculate the sum of divisors for all numbers [1, n]
for i in range(1, n + 1):
for j in range(i * 2, n + 1, i):
dp[j] += i
# Find all abundant numbers up to n
for i in range(12, n + 1):
if dp[i] > i:
abundantNumbers.append(i)
return abundantNumbers
# Driver program
# Check if 12 is an abundant number
if isAbundantNumber(12):
print("YES")
else:
print("NO")
using System;
using System.Collections.Generic;
class Program
{
// Returns true if the given number is Abundant
static bool IsAbundantNumber(int n)
{
// To store the sum of divisors
int sum = 1;
// Loop through the numbers [2,sqrt(n)]
for (int i = 2; i * i <= n; i++)
{
if (n % i == 0)
{
if (i * i != n)
{
sum += i + n / i;
}
else
{
sum += i;
}
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
return sum > n;
}
// Function to find all abundant numbers up to n using DP
static List<int> GetAllAbundantNumbers(int n)
{
List<int> abundantNumbers = new List<int>();
int[] dp = new int[n + 1];
// Calculate the sum of divisors for all numbers [1, n]
for (int i = 1; i <= n; i++)
{
for (int j = i * 2; j <= n; j += i)
{
dp[j] += i;
}
}
// Find all abundant numbers up to n
for (int i = 12; i <= n; i++)
{
if (dp[i] > i)
{
abundantNumbers.Add(i);
}
}
return abundantNumbers;
}
// Driver program
static void Main()
{
// Check if 12 is an abundant number
if (IsAbundantNumber(12))
{
Console.WriteLine("YES");
}
else
{
Console.WriteLine("NO");
}
}
}
// Returns true if the given number is Abundant
function isAbundantNumber(n) {
// To store the sum of divisors
let sum = 1;
// Loop through the numbers [2,sqrt(n)]
for (let i = 2; i * i <= n; i++) {
if (n % i == 0) {
if (i * i != n) {
sum += i + n / i;
} else {
sum += i;
}
}
}
// A number n is said to be Abundant Number if
// sum of all the proper divisors of the number
// is greater than the value of the number n.
if (sum > n) {
return true;
} else {
return false;
}
}
// Function to find all abundant numbers up to n using DP
function getAllAbundantNumbers(n) {
let abundantNumbers = [];
let dp = new Array(n + 1).fill(0);
// Calculate the sum of divisors for all numbers [1, n]
for (let i = 1; i <= n; i++) {
for (let j = i * 2; j <= n; j += i) {
dp[j] += i;
}
}
// Find all abundant numbers up to n
for (let i = 12; i <= n; i++) {
if (dp[i] > i) {
abundantNumbers.push(i);
}
}
return abundantNumbers;
}
// Driver program
// Check if 12 is an abundant number
if (isAbundantNumber(12)) {
console.log("YES");
} else {
console.log("NO");
}
Output
YES
Time Complexity: O(n*sqrt(n)). for a given number n.
Auxiliary Space: O(N)
References:
https://en.wikipedia.org/wiki/Abundant_number
-"
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