Sum and Product of digits in a number that divide the number
Last Updated :
25 Jul, 2022
Given a positive integer N. The task is to find the sum and product of digits of the number which evenly divides the number n.
Examples:
Input: N = 12
Output: Sum = 3, product = 2
1 and 2 divide 12. So, their sum is 3 and product is 2.
Input: N = 1012
Output: Sum = 4, product = 2
1, 1 and 2 divide 1012.
Approach: The idea is to find each digit of the number n by modulus 10 and then check whether it divides n or not. Accordingly, add it to the sum and multiply it with the product. Notice that the digit can be 0, so take care of that case.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void countDigit( int n)
{
int temp = n, sum = 0, product = 1;
while (temp != 0) {
int d = temp % 10;
temp /= 10;
if (d > 0 && n % d == 0) {
sum += d;
product *= d;
}
}
cout << "Sum = " << sum;
cout << "\nProduct = " << product;
}
int main()
{
int n = 1012;
countDigit(n);
return 0;
}
|
Java
import java.lang.*;
import java.util.*;
class GFG
{
static void countDigit( int n)
{
int temp = n, sum = 0 , product = 1 ;
while (temp != 0 )
{
int d = temp % 10 ;
temp /= 10 ;
if (d > 0 && n % d == 0 )
{
sum += d;
product *= d;
}
}
System.out.print( "Sum = " + sum);
System.out.print( "\nProduct = " + product);
}
public static void main(String args[])
{
int n = 1012 ;
countDigit(n);
}
}
|
Python3
def countDigit(n):
temp = n
sum = 0
product = 1
while (temp ! = 0 ):
d = temp % 10
temp / / = 10
if (d > 0 and n % d = = 0 ):
sum + = d
product * = d
print ( "Sum =" , sum )
print ( "Product =" , product)
if __name__ = = '__main__' :
n = 1012
countDigit(n)
|
C#
using System;
class GFG
{
static void countDigit( int n)
{
int temp = n, sum = 0, product = 1;
while (temp != 0)
{
int d = temp % 10;
temp /= 10;
if (d > 0 && n % d == 0)
{
sum += d;
product *= d;
}
}
Console.Write( "Sum = " + sum);
Console.Write( "\nProduct = " + product);
}
public static void Main()
{
int n = 1012;
countDigit(n);
}
}
|
PHP
<?php
function countDigit( $n )
{
$temp = $n ;
$sum = 0;
$product = 1;
while ( $temp != 0) {
$d = $temp % 10;
$temp =(int)( $temp /10);
if ( $d > 0 && $n % $d == 0) {
$sum += $d ;
$product *= $d ;
}
}
echo "Sum = " . $sum ;
echo "\nProduct = " . $product ;
}
$n = 1012;
countDigit( $n );
?>
|
Javascript
<script>
function countDigit(n)
{
let temp = n;
let sum = 0;
let product = 1;
while (temp != 0) {
let d = temp % 10;
temp =parseInt(temp/10);
if (d > 0 && n % d == 0) {
sum += d;
product *= d;
}
}
document.write( "Sum = " +sum);
document.write( "<br>Product = " +product);
}
let n = 1012;
countDigit(n);
</script>
|
Output:
Sum = 4
Product = 2
Time Complexity: O(log10n), Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...