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# Multiply perfect number

• Last Updated : 16 Jul, 2021

A number N is said to be Multiply-perfect numbers if N divides sigma(N), where sigma(N) = sum of all divisors of N.
The first few Multiply-perfect numbers are:

1, 6, 28, 120, 496, 672, ……..

### Check if N is a Multiply-perfect number

Given a number N, the task is to find if this number is Multiply-perfect number or not.
Examples:

Input: N = 120
Output: YES
Explanation:
Sum of 120’s divisors is 1 + 2 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 + 120 = 360 and 120 divides 360.
Therefore, 120 is a Multiply-perfect number.
Input: N = 32
Output: No

Approach: For a number N to be Multiply-perfect number, the following condition should hold true: sigma(N) % N = 0, where sigma(N) = sum of all divisors of N. Therefore, we will find sum of all divisors of N and check if it is divisible by N or not. If divisible print “Yes” else print “No.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;` `// Function to find the``// 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;` `            ``// Otherwise take both``            ``else` `{``                ``sum = sum + i;``                ``sum = sum + (n / i);``            ``}``        ``}``    ``}` `    ``return` `sum;``}` `// Function to check Multiply-perfect number``bool` `MultiplyPerfectNumber(``int` `n)``{``    ``if` `(getSum(n) % n == 0)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Driver code``int` `main()``{` `    ``int` `n = 28;``    ``if` `(MultiplyPerfectNumber(n)) {``        ``cout << ``"Yes"``;``    ``}``    ``else` `{``        ``cout << ``"No"``;``    ``}``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG{``    ` `// Function to find the``// 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 Multiply-perfect number``static` `boolean` `MultiplyPerfectNumber(``int` `n)``{``    ``if` `(getSum(n) % n == ``0``)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``28``;``    ` `    ``if` `(MultiplyPerfectNumber(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 find the``# sum of divisors``def` `getSum(n):` `    ``sum1 ``=` `0``;` `    ``# Note that this loop``    ``# runs till square root of N``    ``for` `i ``in` `range``(``1``, ``int``(math.sqrt(n))):``        ``if` `(n ``%` `i ``=``=` `0``):` `            ``# If divisors are equal,``            ``# take only one of them``            ``if` `(n ``/``/` `i ``=``=` `i):``                ``sum1 ``=` `sum1 ``+` `i;` `            ``# Otherwise take both``            ``else``:``                ``sum1 ``=` `sum1 ``+` `i;``                ``sum1 ``=` `sum1 ``+` `(n ``/``/` `i);``            ` `    ``return` `sum1;` `# Function to check Multiply-perfect number``def` `MultiplyPerfectNumber(n):` `    ``if` `(getSum(n) ``%` `n ``=``=` `0``):``        ``return` `True``;``    ``else``:``        ``return` `False``;` `# Driver code``n ``=` `28``;``if` `(MultiplyPerfectNumber(n)):``    ``print``(``"Yes"``);``else``:``    ``print``(``"No"``);` `# This code is contributed by Code_Mech`

## C#

 `// C# implementation of the above approach``using` `System;``class` `GFG{``    ` `// Function to find the``// 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 Multiply-perfect number``static` `bool` `MultiplyPerfectNumber(``int` `n)``{``    ``if` `(getSum(n) % n == 0)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 28;``    ` `    ``if` `(MultiplyPerfectNumber(n))``    ``{``        ``Console.Write(``"Yes"``);``    ``}``    ``else``    ``{``        ``Console.Write(``"No"``);``    ``}``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``
Output:
`Yes`

Time Complexity: O(N1/2)

Auxiliary Space: O(1)

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