# Arithmetic Number

In number theory, an arithmetic number is an integer for which the average of its positive divisors is also an integer. Or in other words, a number N is arithmetic if the number of divisors divides the sum of divisors.
Given a positive integer n. The task is to check whether n is Arithmetic number or not.

Examples:

```Input : n = 6
Output : Yes
Sum of divisor of 6 = 1 + 2 + 3 + 6 = 12.
Number of divisor of 6 = 4.
So, on dividing Sum of divisor by Number of divisor
= 12/4 = 3, which is an integer.

Input : n = 2
Output : No
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Algorithm

1. Find sum of all the factors of a number , say sum.
2. Find count of divisors (say Count).
3. Check if sum is divisible by count.

## C++

 `// CPP program to check if a number is Arithmetic ` `// number or not ` `#include ` `using` `namespace` `std; ` ` `  `// Sieve Of Eratosthenes ` `void` `SieveOfEratosthenes(``int` `n, ``bool` `prime[], ` `                         ``bool` `primesquare[], ``int` `a[]) ` `{ ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``prime[i] = ``true``; ` ` `  `    ``for` `(``int` `i = 0; i <= (n * n + 1); i++) ` `        ``primesquare[i] = ``false``; ` ` `  `    ``// 1 is not a prime number ` `    ``prime = ``false``; ` ` `  `    ``for` `(``int` `p = 2; p * p <= n; p++) { ` `        ``// If prime[p] is not changed, then ` `        ``// it is a prime ` `        ``if` `(prime[p] == ``true``) { ` `            ``// Update all multiples of p ` `            ``for` `(``int` `i = p * 2; i <= n; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` ` `  `    ``int` `j = 0; ` `    ``for` `(``int` `p = 2; p <= n; p++) { ` `        ``if` `(prime[p]) { ` `            ``// Storing primes in an array ` `            ``a[j] = p; ` ` `  `            ``// Update value in primesquare[p*p], ` `            ``// if p is prime. ` `            ``primesquare[p * p] = ``true``; ` `            ``j++; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to count divisors ` `int` `countDivisors(``int` `n) ` `{ ` `    ``// If number is 1, then it will have only 1 ` `    ``// as a factor. So, total factors will be 1. ` `    ``if` `(n == 1) ` `        ``return` `1; ` ` `  `    ``bool` `prime[n + 1], primesquare[n * n + 1]; ` ` `  `    ``int` `a[n]; ``// for storing primes upto n ` ` `  `    ``// Calling SieveOfEratosthenes to store prime ` `    ``// factors of n and to store square of prime ` `    ``// factors of n ` `    ``SieveOfEratosthenes(n, prime, primesquare, a); ` ` `  `    ``// ans will contain total number of  ` `    ``// distinct divisors ` `    ``int` `ans = 1; ` ` `  `    ``// Loop for counting factors of n ` `    ``for` `(``int` `i = 0;; i++) { ` ` `  `        ``// a[i] is not less than cube root n ` `        ``if` `(a[i] * a[i] * a[i] > n) ` `            ``break``; ` ` `  `        ``// Calculating power of a[i] in n. ` `        ``// cnt is power of prime a[i] in n. ` `        ``int` `cnt = 1;  ` ` `  `        ``// if a[i] is a factor of n ` `        ``while` `(n % a[i] == 0)  ` `        ``{ ` `            ``n = n / a[i]; ` `            ``cnt = cnt + 1; ``// incrementing power ` `        ``} ` ` `  `        ``// Calculating number of divisors ` `        ``// If n = a^p * b^q then total  ` `        ``// divisors of n ` `        ``// are (p+1)*(q+1) ` `        ``ans = ans * cnt; ` `    ``} ` ` `  `    ``// if a[i] is greater than cube root of n ` ` `  `    ``// First case ` `    ``if` `(prime[n]) ` `        ``ans = ans * 2; ` ` `  `    ``// Second case ` `    ``else` `if` `(primesquare[n]) ` `        ``ans = ans * 3; ` ` `  `    ``// Third casse ` `    ``else` `if` `(n != 1) ` `        ``ans = ans * 4; ` ` `  `    ``return` `ans; ``// Total divisors ` `} ` ` `  `// Returns sum of all factors of n. ` `int` `sumofFactors(``int` `n) ` `{ ` `    ``// Traversing through all prime factors. ` `    ``int` `res = 1; ` `    ``for` `(``int` `i = 2; i <= ``sqrt``(n); i++) { ` ` `  `        ``int` `count = 0, curr_sum = 1; ` `        ``int` `curr_term = 1; ` `        ``while` `(n % i == 0) { ` `            ``count++; ` `            ``n = n / i; ` ` `  `            ``curr_term *= i; ` `            ``curr_sum += curr_term; ` `        ``} ` ` `  `        ``res *= curr_sum; ` `    ``} ` ` `  `    ``// This condition is to handle ` `    ``// the case when n is a prime ` `    ``// number greater than 2. ` `    ``if` `(n >= 2) ` `        ``res *= (1 + n); ` ` `  `    ``return` `res; ` `} ` ` `  `// Check if number is Arithmetic Number ` `// or not. ` `bool` `checkArithmetic(``int` `n) ` `{ ` `    ``int` `count = countDivisors(n); ` `    ``int` `sum = sumofFactors(n); ` ` `  `    ``return` `(sum  % count == 0); ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `n = 6; ` `    ``(checkArithmetic(n)) ? (cout << ``"Yes"``) :  ` `                           ``(cout << ``"No"``); ` `    ``return` `0; ` `} `

## Java

 `// Java program to check if a number is Arithmetic ` `// number or not ` `class` `GFG  ` `{ ` `     `  `// Sieve Of Eratosthenes ` `static` `void` `SieveOfEratosthenes(``int` `n, ``boolean` `prime[], ` `                        ``boolean` `primesquare[], ``int` `a[]) ` `{ ` `    ``for` `(``int` `i = ``2``; i <= n; i++) ` `        ``prime[i] = ``true``; ` ` `  `    ``for` `(``int` `i = ``0``; i <= (n * n ); i++) ` `        ``primesquare[i] = ``false``; ` ` `  `    ``// 1 is not a prime number ` `    ``prime[``1``] = ``false``; ` ` `  `    ``for` `(``int` `p = ``2``; p * p <= n; p++)  ` `    ``{ ` `        ``// If prime[p] is not changed, then ` `        ``// it is a prime ` `        ``if` `(prime[p] == ``true``)  ` `        ``{ ` `            ``// Update all multiples of p ` `            ``for` `(``int` `i = p * ``2``; i <= n; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` ` `  `    ``int` `j = ``0``; ` `    ``for` `(``int` `p = ``2``; p <= n; p++)  ` `    ``{ ` `        ``if` `(prime[p])  ` `        ``{ ` `            ``// Storing primes in an array ` `            ``a[j] = p; ` ` `  `            ``// Update value in primesquare[p*p], ` `            ``// if p is prime. ` `            ``primesquare[p * p] = ``true``; ` `            ``j++; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to count divisors ` `static` `int` `countDivisors(``int` `n) ` `{ ` `    ``// If number is 1, then it will have only 1 ` `    ``// as a factor. So, total factors will be 1. ` `    ``if` `(n == ``1``) ` `        ``return` `1``; ` ` `  `    ``boolean` `prime[] = ``new` `boolean``[n + ``1``], ` `            ``primesquare[] = ``new` `boolean``[n * n + ``1``]; ` ` `  `    ``int` `a[] = ``new` `int``[n]; ``// for storing primes upto n ` ` `  `    ``// Calling SieveOfEratosthenes to store prime ` `    ``// factors of n and to store square of prime ` `    ``// factors of n ` `    ``SieveOfEratosthenes(n, prime, primesquare, a); ` ` `  `    ``// ans will contain total number of  ` `    ``// distinct divisors ` `    ``int` `ans = ``1``; ` ` `  `    ``// Loop for counting factors of n ` `    ``for` `(``int` `i = ``0``;; i++)  ` `    ``{ ` ` `  `        ``// a[i] is not less than cube root n ` `        ``if` `(a[i] * a[i] * a[i] > n) ` `            ``break``; ` ` `  `        ``// Calculating power of a[i] in n. ` `        ``// cnt is power of prime a[i] in n. ` `        ``int` `cnt = ``1``;  ` ` `  `        ``// if a[i] is a factor of n ` `        ``while` `(n % a[i] == ``0``)  ` `        ``{ ` `            ``n = n / a[i]; ` `            ``cnt = cnt + ``1``; ``// incrementing power ` `        ``} ` ` `  `        ``// Calculating number of divisors ` `        ``// If n = a^p * b^q then total  ` `        ``// divisors of n ` `        ``// are (p+1)*(q+1) ` `        ``ans = ans * cnt; ` `    ``} ` ` `  `    ``// if a[i] is greater than cube root of n ` ` `  `    ``// First case ` `    ``if` `(prime[n]) ` `        ``ans = ans * ``2``; ` ` `  `    ``// Second case ` `    ``else` `if` `(primesquare[n]) ` `        ``ans = ans * ``3``; ` ` `  `    ``// Third casse ` `    ``else` `if` `(n != ``1``) ` `        ``ans = ans * ``4``; ` ` `  `    ``return` `ans; ``// Total divisors ` `} ` ` `  `// Returns sum of all factors of n. ` `static` `int` `sumofFactors(``int` `n) ` `{ ` `    ``// Traversing through all prime factors. ` `    ``int` `res = ``1``; ` `    ``for` `(``int` `i = ``2``; i <= Math.sqrt(n); i++)  ` `    ``{ ` ` `  `        ``int` `count = ``0``, curr_sum = ``1``; ` `        ``int` `curr_term = ``1``; ` `        ``while` `(n % i == ``0``)  ` `        ``{ ` `            ``count++; ` `            ``n = n / i; ` ` `  `            ``curr_term *= i; ` `            ``curr_sum += curr_term; ` `        ``} ` ` `  `        ``res *= curr_sum; ` `    ``} ` ` `  `    ``// This condition is to handle ` `    ``// the case when n is a prime ` `    ``// number greater than 2. ` `    ``if` `(n >= ``2``) ` `        ``res *= (``1` `+ n); ` ` `  `    ``return` `res; ` `} ` ` `  `// Check if number is Arithmetic Number ` `// or not. ` `static` `boolean` `checkArithmetic(``int` `n) ` `{ ` `    ``int` `count = countDivisors(n); ` `    ``int` `sum = sumofFactors(n); ` ` `  `    ``return` `(sum % count == ``0``); ` `} ` ` `  `// Driver Program ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `n = ``6``; ` `    ``if``(checkArithmetic(n)) ` `        ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python3 program to check if  ` `# a number is Arithmetic  ` `# number or not ` `import` `math ` ` `  `# Sieve Of Eratosthenes ` `def` `SieveOfEratosthenes(n, prime,primesquare, a): ` ` `  `    ``for` `i ``in` `range``(``2``,n``+``1``): ` `        ``prime[i] ``=` `True``; ` ` `  `    ``for` `i ``in` `range``((n ``*` `n ``+` `1``)``+``1``): ` `        ``primesquare[i] ``=` `False``; ` ` `  `    ``# 1 is not a ` `    ``# prime number ` `    ``prime[``1``] ``=` `False``; ` `    ``p ``=` `2``; ` ` `  `    ``while` `(p ``*` `p <``=` `n): ` `         `  `        ``# If prime[p] is  ` `        ``# not changed, then ` `        ``# it is a prime ` `        ``if` `(prime[p] ``=``=` `True``): ` `            ``# Update all multiples of p ` `            ``for` `i ``in` `range``(p ``*` `2``,n``+``1``,p): ` `                ``prime[i] ``=` `False``; ` `        ``p``+``=``1``; ` ` `  `    ``j ``=` `0``; ` `    ``for` `p ``in` `range``(``2``,n``+``1``): ` `        ``if` `(prime[p]):  ` `             `  `            ``# Storing primes in an array ` `            ``a[j] ``=` `p; ` ` `  `            ``# Update value in ` `            ``# primesquare[p*p], ` `            ``# if p is prime. ` `            ``primesquare[p ``*` `p] ``=` `True``; ` `            ``j``+``=``1``; ` ` `  `# Function to count divisors ` `def` `countDivisors(n): ` ` `  `    ``# If number is 1, then it ` `    ``# will have only 1 as a  ` `    ``# factor. So, total factors  ` `    ``# will be 1. ` `    ``if` `(n ``=``=` `1``): ` `        ``return` `1``; ` ` `  `    ``prime ``=` `[``False``]``*``(n ``+` `2``); ` `    ``primesquare ``=` `[``False``]``*``(n ``*``n ``+` `3``); ` ` `  `    ``# for storing primes upto n ` `    ``a ``=` `[``0``]``*``n;  ` ` `  `    ``# Calling SieveOfEratosthenes  ` `    ``# to store prime factors of  ` `    ``# n and to store square of  ` `    ``# prime factors of n ` `    ``SieveOfEratosthenes(n, prime,primesquare, a); ` ` `  `    ``# ans will contain  ` `    ``# total number of  ` `    ``# distinct divisors ` `    ``ans ``=` `1``; ` ` `  `    ``# Loop for counting ` `    ``# factors of n ` `    ``for` `i ``in` `range``(``0``,``True``): ` ` `  `        ``# a[i] is not less  ` `        ``# than cube root n ` `        ``if` `(a[i] ``*` `a[i] ``*` `a[i] > n): ` `            ``break``; ` ` `  `        ``# Calculating power of  ` `        ``# a[i] in n. cnt is power  ` `        ``# of prime a[i] in n. ` `        ``cnt ``=` `1``;  ` ` `  `        ``# if a[i] is a factor of n ` `        ``while` `(n ``%` `a[i] ``=``=` `0``):  ` `            ``n ``/``/``=` `a[i]; ` `             `  `            ``# incrementing power ` `            ``cnt ``=` `cnt ``+` `1``;  ` ` `  `        ``# Calculating number of  ` `        ``# divisors. If n = a^p * b^q  ` `        ``# then total divisors  ` `        ``# of n are (p+1)*(q+1) ` `        ``ans ``=` `ans ``*` `cnt; ` ` `  `    ``# if a[i] is greater  ` `    ``# than cube root of n ` ` `  `    ``# First case ` `    ``if` `(prime[n]): ` `        ``ans ``=` `ans ``*` `2``; ` ` `  `    ``# Second case ` `    ``elif` `(primesquare[n]): ` `        ``ans ``=` `ans ``*` `3``; ` ` `  `    ``# Third casse ` `    ``elif` `(n !``=` `1``): ` `        ``ans ``=` `ans ``*` `4``; ` ` `  `    ``return` `ans; ``# Total divisors ` ` `  `# Returns sum of ` `# all factors of n. ` `def` `sumofFactors(n): ` ` `  `    ``# Traversing through  ` `    ``# all prime factors. ` `    ``res ``=` `1``; ` `    ``for` `i ``in` `range``(``2``,``int``(math.sqrt(n))``+``1``):  ` `        ``count ``=` `0``;  ` `        ``curr_sum ``=` `1``; ` `        ``curr_term ``=` `1``; ` `        ``while` `(n ``%` `i ``=``=` `0``): ` `            ``count``+``=``1``; ` `            ``n ``/``/``=` `i; ` ` `  `            ``curr_term ``*``=` `i; ` `            ``curr_sum ``+``=` `curr_term; ` ` `  `        ``res ``*``=` `curr_sum; ` ` `  `    ``# This condition is to handle ` `    ``# the case when n is a prime ` `    ``# number greater than 2. ` `    ``if` `(n >``=` `2``): ` `        ``res ``*``=` `(``1` `+` `n); ` ` `  `    ``return` `res; ` ` `  `# Check if number is  ` `# Arithmetic Number or not. ` `def` `checkArithmetic(n): ` ` `  `    ``count ``=` `countDivisors(n); ` `    ``sum` `=` `sumofFactors(n); ` ` `  `    ``return` `(``sum` `%` `count ``=``=` `0``); ` ` `  `# Driver code ` `n ``=` `6``; ` `if``(checkArithmetic(n)): ` `    ``print``(``"Yes"``); ` `else``: ` `    ``print``(``"No"``); ` ` `  `# This code is contributed ` `# by mits `

## C#

 `// C# program to check if a number  ` `// is arithmetic number or not ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `// Sieve Of Eratosthenes ` `static` `void` `SieveOfEratosthenes(``int` `n, ``bool` `[]prime, ` `                        ``bool` `[]primesquare, ``int` `[]a) ` `{ ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``prime[i] = ``true``; ` ` `  `    ``for` `(``int` `i = 0; i <= (n * n ); i++) ` `        ``primesquare[i] = ``false``; ` ` `  `    ``// 1 is not a prime number ` `    ``prime = ``false``; ` ` `  `    ``for` `(``int` `p = 2; p * p <= n; p++)  ` `    ``{ ` `        ``// If prime[p] is not changed, then ` `        ``// it is a prime ` `        ``if` `(prime[p] == ``true``)  ` `        ``{ ` `            ``// Update all multiples of p ` `            ``for` `(``int` `i = p * 2; i <= n; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` ` `  `    ``int` `j = 0; ` `    ``for` `(``int` `p = 2; p <= n; p++)  ` `    ``{ ` `        ``if` `(prime[p])  ` `        ``{ ` `            ``// Storing primes in an array ` `            ``a[j] = p; ` ` `  `            ``// Update value in primesquare[p*p], ` `            ``// if p is prime. ` `            ``primesquare[p * p] = ``true``; ` `            ``j++; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to count divisors ` `static` `int` `countDivisors(``int` `n) ` `{ ` `    ``// If number is 1, then it will have only 1 ` `    ``// as a factor. So, total factors will be 1. ` `    ``if` `(n == 1) ` `        ``return` `1; ` ` `  `    ``bool` `[]prime = ``new` `bool``[n + 1]; ` `    ``bool` `[]primesquare = ``new` `bool``[n * n + 1]; ` ` `  `    ``int` `[]a = ``new` `int``[n]; ``// for storing primes upto n ` ` `  `    ``// Calling SieveOfEratosthenes to store prime ` `    ``// factors of n and to store square of prime ` `    ``// factors of n ` `    ``SieveOfEratosthenes(n, prime, primesquare, a); ` ` `  `    ``// ans will contain total number of  ` `    ``// distinct divisors ` `    ``int` `ans = 1; ` ` `  `    ``// Loop for counting factors of n ` `    ``for` `(``int` `i = 0;; i++)  ` `    ``{ ` ` `  `        ``// a[i] is not less than cube root n ` `        ``if` `(a[i] * a[i] * a[i] > n) ` `            ``break``; ` ` `  `        ``// Calculating power of a[i] in n. ` `        ``// cnt is power of prime a[i] in n. ` `        ``int` `cnt = 1;  ` ` `  `        ``// if a[i] is a factor of n ` `        ``while` `(n % a[i] == 0)  ` `        ``{ ` `            ``n = n / a[i]; ` `            ``cnt = cnt + 1; ``// incrementing power ` `        ``} ` ` `  `        ``// Calculating number of divisors ` `        ``// If n = a^p * b^q then total  ` `        ``// divisors of n ` `        ``// are (p+1)*(q+1) ` `        ``ans = ans * cnt; ` `    ``} ` ` `  `    ``// if a[i] is greater than cube root of n ` ` `  `    ``// First case ` `    ``if` `(prime[n]) ` `        ``ans = ans * 2; ` ` `  `    ``// Second case ` `    ``else` `if` `(primesquare[n]) ` `        ``ans = ans * 3; ` ` `  `    ``// Third casse ` `    ``else` `if` `(n != 1) ` `        ``ans = ans * 4; ` ` `  `    ``return` `ans; ``// Total divisors ` `} ` ` `  `// Returns sum of all factors of n. ` `static` `int` `sumofFactors(``int` `n) ` `{ ` `    ``// Traversing through all prime factors. ` `    ``int` `res = 1; ` `    ``for` `(``int` `i = 2; i <= Math.Sqrt(n); i++)  ` `    ``{ ` ` `  `        ``int` `count = 0, curr_sum = 1; ` `        ``int` `curr_term = 1; ` `        ``while` `(n % i == 0)  ` `        ``{ ` `            ``count++; ` `            ``n = n / i; ` ` `  `            ``curr_term *= i; ` `            ``curr_sum += curr_term; ` `        ``} ` ` `  `        ``res *= curr_sum; ` `    ``} ` ` `  `    ``// This condition is to handle ` `    ``// the case when n is a prime ` `    ``// number greater than 2. ` `    ``if` `(n >= 2) ` `        ``res *= (1 + n); ` ` `  `    ``return` `res; ` `} ` ` `  `// Check if number is Arithmetic Number ` `// or not. ` `static` `bool` `checkArithmetic(``int` `n) ` `{ ` `    ``int` `count = countDivisors(n); ` `    ``int` `sum = sumofFactors(n); ` ` `  `    ``return` `(sum % count == 0); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `n = 6; ` `    ``if``(checkArithmetic(n)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## PHP

 ` ``\$n``) ` `            ``break``; ` ` `  `        ``// Calculating power of  ` `        ``// a[i] in n. cnt is power  ` `        ``// of prime a[i] in n. ` `        ``\$cnt` `= 1;  ` ` `  `        ``// if a[i] is a factor of n ` `        ``while` `(``\$n` `% ``\$a``[``\$i``] == 0)  ` `        ``{ ` `            ``\$n` `= (int)(``\$n` `/ ``\$a``[``\$i``]); ` `             `  `            ``// incrementing power ` `            ``\$cnt` `= ``\$cnt` `+ 1;  ` `        ``} ` ` `  `        ``// Calculating number of  ` `        ``// divisors. If n = a^p * b^q  ` `        ``// then total divisors  ` `        ``// of n are (p+1)*(q+1) ` `        ``\$ans` `= ``\$ans` `* ``\$cnt``; ` `    ``} ` ` `  `    ``// if a[i] is greater  ` `    ``// than cube root of n ` ` `  `    ``// First case ` `    ``if` `(``\$prime``[``\$n``]) ` `        ``\$ans` `= ``\$ans` `* 2; ` ` `  `    ``// Second case ` `    ``else` `if` `(``\$primesquare``[``\$n``]) ` `        ``\$ans` `= ``\$ans` `* 3; ` ` `  `    ``// Third casse ` `    ``else` `if` `(``\$n` `!= 1) ` `        ``\$ans` `= ``\$ans` `* 4; ` ` `  `    ``return` `\$ans``; ``// Total divisors ` `} ` ` `  `// Returns sum of ` `// all factors of n. ` `function` `sumofFactors(``\$n``) ` `{ ` `    ``// Traversing through  ` `    ``// all prime factors. ` `    ``\$res` `= 1; ` `    ``for` `(``\$i` `= 2;  ` `         ``\$i` `<= sqrt(``\$n``); ``\$i``++)  ` `    ``{ ` `        ``\$count` `= 0;  ` `        ``\$curr_sum` `= 1; ` `        ``\$curr_term` `= 1; ` `        ``while` `(``\$n` `% ``\$i` `== 0)  ` `        ``{ ` `            ``\$count``++; ` `            ``\$n` `= (int)(``\$n` `/ ``\$i``); ` ` `  `            ``\$curr_term` `*= ``\$i``; ` `            ``\$curr_sum` `+= ``\$curr_term``; ` `        ``} ` ` `  `        ``\$res` `*= ``\$curr_sum``; ` `    ``} ` ` `  `    ``// This condition is to handle ` `    ``// the case when n is a prime ` `    ``// number greater than 2. ` `    ``if` `(``\$n` `>= 2) ` `        ``\$res` `*= (1 + ``\$n``); ` ` `  `    ``return` `\$res``; ` `} ` ` `  `// Check if number is  ` `// Arithmetic Number or not. ` `function` `checkArithmetic(``\$n``) ` `{ ` `    ``\$count` `= countDivisors(``\$n``); ` `    ``\$sum` `= sumofFactors(``\$n``); ` ` `  `    ``return` `(``\$sum` `% ``\$count` `== 0); ` `} ` ` `  `// Driver code ` `\$n` `= 6; ` `echo` `(checkArithmetic(``\$n``)) ?  ` `                     ``"Yes"` `: ``"No"``; ` ` `  `// This code is contributed ` `// by mits ` `?> `

Output:

```Yes
```

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