Find sum of odd factors of a number
Given a number n, the task is to find the odd factor sum.
Examples :
Input : n = 30 Output : 24 Odd dividers sum 1 + 3 + 5 + 15 = 24 Input : 18 Output : 13 Odd dividers sum 1 + 3 + 9 = 13
Prerequisite : Sum of all the factors of a number
As discussed in above mentioned previous post, sum of factors of a number is
Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).
Sum of divisors = (1 + p1 + p12 ... p1a1) *
(1 + p2 + p22 ... p2a2) *
.............................................
(1 + pk + pk2 ... pkak)
To find sum of odd factors, we simply need to ignore even factors and their powers. For example, consider n = 18. It can be written as 2132 and sun of all factors is (1)*(1 + 2)*(1 + 3 + 32). Sum of odd factors (1)*(1+3+32) = 13.
To remove all even factors, we repeatedly divide n while it is divisible by 2. After this step, we only get odd factors. Note that 2 is the only even prime.
C++
// Formula based CPP program // to find sum of all // divisors of n. #include <bits/stdc++.h> using namespace std; // Returns sum of all factors of n. int sumofoddFactors(int n) { // Traversing through all // prime factors. int res = 1; // ignore even factors by // removing all powers of // 2 while (n % 2 == 0) n = n / 2; for (int i = 3; i <= sqrt(n); i++) { // While i divides n, print // i and divide n 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. if (n >= 2) res *= (1 + n); return res; } // Driver code int main() { int n = 30; cout << sumofoddFactors(n); return 0; } |
Java
// Formula based Java program // to find sum of all divisors // of n. import java.io.*; import java.math.*; class GFG { // Returns sum of all // factors of n. static int sumofoddFactors(int n) { // Traversing through // all prime factors. int res = 1; // ignore even factors by // removing all powers // of 2 while (n % 2 == 0) n = n / 2; for (int i = 3; i <= Math.sqrt(n); i++) { // While i divides n, print i // and divide n 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. if (n >= 2) res *= (1 + n); return res; } // Driver code public static void main(String args[]) throws IOException { int n = 30; System.out.println(sumofoddFactors(n)); } } /* This code is contributed by Nikita Tiwari.*/ |
Python3
# Formula based Python3 program # to find sum of all divisors # of n. import math # Returns sum of all factors # of n. def sumofoddFactors( n ): # Traversing through all # prime factors. res = 1 # ignore even factors by # of 2 while n % 2 == 0: n = n // 2 for i in range(3, int(math.sqrt(n) + 1)): # While i divides n, print # i and divide n count = 0 curr_sum = 1 curr_term = 1 while n % i == 0: count+=1 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. if n >= 2: res *= (1 + n) return res # Driver code n = 30print(sumofoddFactors(n)) # This code is contributed by "Sharad_Bhardwaj". |
C#
// Formula based C# program to // find sum of all divisors of n. using System; class GFG { // Returns sum of all // factors of n. static int sumofoddFactors(int n) { // Traversing through // all prime factors. int res = 1; // ignore even factors by // removing all powers // of 2 while (n % 2 == 0) n = n / 2; for (int i = 3; i <= Math.Sqrt(n); i++) { // While i divides n, print i // and divide n 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. if (n >= 2) res *= (1 + n); return res; } // Driver code public static void Main(String[] argc) { int n = 30; Console.Write(sumofoddFactors(n)); } } /* This code is contributed by parashar...*/ |
PHP
<?php // Formula based PHP program // to find sum of all // divisors of n. // Returns sum of all factors of n. function sumofoddFactors($n) { // Traversing through all // prime factors. $res = 1; // ignore even factors by // removing all powers of // 2 while ($n % 2 == 0) $n = $n / 2; for ($i = 3; $i <= sqrt($n); $i++) { // While i divides n, print // i and divide n $count = 0; $curr_sum = 1; $curr_term = 1; while ($n % $i == 0) { $count++; $n = $n / $i; $curr_term *= $i; $curr_sum += $curr_term; } $res *= $curr_sum; } // This condition is to // handle the case when // n is a prime number. if ($n >= 2) $res *= (1 + $n); return $res; } // Driver code $n = 30; echo sumofoddFactors($n); // This code is contributed // by nitin mittal. ?> |
Output :
24
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