Given n and a number, the task is to find the sum of n digit numbers that are divisible by given number.
Examples:
Input : n = 2, number = 7
Output : 728
Explanation:
There are thirteen n digit numbers that are divisible by 7.
Numbers are : 14+ 21 + 28 + 35 + 42 + 49 + 56 + 63 +70 + 77 + 84 + 91 + 98.Input : n = 3, number = 7
Output : 70336Input : n = 3, number = 4
Output : 123300
Native Approach: Traverse through all n digit numbers. For every number check for divisibility, and make the sum.
// Simple CPP program to sum of n digit // divisible numbers. #include <cmath> #include <iostream> using namespace std;
// Returns sum of n digit numbers // divisible by 'number' int totalSumDivisibleByNum( int n, int number)
{ // compute the first and last term
int firstnum = pow (10, n - 1);
int lastnum = pow (10, n);
// sum of number which having
// n digit and divisible by number
int sum = 0;
for ( int i = firstnum; i < lastnum; i++)
if (i % number == 0)
sum += i;
return sum;
} // Driver code int main()
{ int n = 3, num = 7;
cout << totalSumDivisibleByNum(n, num) << "\n" ;
return 0;
} |
// Simple Java program to sum of n digit // divisible numbers. import java.io.*;
class GFG {
// Returns sum of n digit numbers
// divisible by 'number'
static int totalSumDivisibleByNum( int n, int number)
{
// compute the first and last term
int firstnum = ( int )Math.pow( 10 , n - 1 );
int lastnum = ( int )Math.pow( 10 , n);
// sum of number which having
// n digit and divisible by number
int sum = 0 ;
for ( int i = firstnum; i < lastnum; i++)
if (i % number == 0 )
sum += i;
return sum;
}
// Driver code
public static void main (String[] args)
{
int n = 3 , num = 7 ;
System.out.println(totalSumDivisibleByNum(n, num));
}
} // This code is contributed by Ajit. |
# Simple Python 3 program to sum # of n digit divisible numbers. # Returns sum of n digit numbers # divisible by 'number' def totalSumDivisibleByNum(n, number):
# compute the first and last term
firstnum = pow ( 10 , n - 1 )
lastnum = pow ( 10 , n)
# sum of number which having
# n digit and divisible by number
sum = 0
for i in range (firstnum, lastnum):
if (i % number = = 0 ):
sum + = i
return sum
# Driver code n = 3 ; num = 7
print (totalSumDivisibleByNum(n, num))
# This code is contributed by Smitha Dinesh Semwal |
// Simple C# program to sum of n digit // divisible numbers. using System;
class GFG {
// Returns sum of n digit numbers
// divisible by 'number'
static int totalSumDivisibleByNum( int n, int number)
{
// compute the first and last term
int firstnum = ( int )Math.Pow(10, n - 1);
int lastnum = ( int )Math.Pow(10, n);
// sum of number which having
// n digit and divisible by number
int sum = 0;
for ( int i = firstnum; i < lastnum; i++)
if (i % number == 0)
sum += i;
return sum;
}
// Driver code
public static void Main ()
{
int n = 3, num = 7;
Console.WriteLine(totalSumDivisibleByNum(n, num));
}
} // This code is contributed by vt_m. |
<?php // Simple PHP program to sum of // n digit divisible numbers. // Returns sum of n digit numbers // divisible by 'number' function totalSumDivisibleByNum( $n , $number )
{ // compute the first and last term
$firstnum = pow(10, $n - 1);
$lastnum = pow(10, $n );
// sum of number which having
// n digit and divisible by number
$sum = 0;
for ( $i = $firstnum ; $i < $lastnum ; $i ++)
if ( $i % $number == 0)
$sum += $i ;
return $sum ;
} // Driver code
$n = 3; $num = 7;
echo totalSumDivisibleByNum( $n , $num ) , "\n" ;
// This code is contributed by aj_36 ?> |
<script> // JavaScript program to sum of n digit // divisible numbers. // Returns sum of n digit numbers // divisible by 'number' function totalSumDivisibleByNum(n, number)
{ // compute the first and last term
let firstnum = Math.pow(10, n - 1);
let lastnum = Math.pow(10, n);
// sum of number which having
// n digit and divisible by number
let sum = 0;
for (let i = firstnum; i < lastnum; i++)
if (i % number == 0)
sum += i;
return sum;
} // Driver Code let n = 3, num = 7; document.write(totalSumDivisibleByNum(n, num)); // This code is contributed by chinmoy1997pal </script> |
70336
Time Complexity: O(10n)
Auxiliary Space: O(1)
Efficient Method :
First, find the count of n digit numbers divisible by a given number. Then apply formula for sum of AP.
count/2 * (first-term + last-term)
// Efficient CPP program to find the sum // divisible numbers. #include <cmath> #include <iostream> using namespace std;
// find the Sum of having n digit and // divisible by the number int totalSumDivisibleByNum( int digit,
int number)
{ // compute the first and last term
int firstnum = pow (10, digit - 1);
int lastnum = pow (10, digit);
// first number which is divisible
// by given number
firstnum = (firstnum - firstnum % number)
+ number;
// last number which is divisible
// by given number
lastnum = (lastnum - lastnum % number);
// total divisible number
int count = ((lastnum - firstnum) /
number + 1);
// return the total sum
return ((lastnum + firstnum) * count) / 2;
} int main()
{ int n = 3, number = 7;
cout << totalSumDivisibleByNum(n, number);
return 0;
} |
// Efficient Java program to find the sum // divisible numbers. import java.io.*;
class GFG {
// find the Sum of having n digit and
// divisible by the number
static int totalSumDivisibleByNum( int digit,
int number)
{
// compute the first and last term
int firstnum = ( int )Math.pow( 10 , digit - 1 );
int lastnum = ( int )Math.pow( 10 , digit);
// first number which is divisible
// by given number
firstnum = (firstnum - firstnum % number)
+ number;
// last number which is divisible
// by given number
lastnum = (lastnum - lastnum % number);
// total divisible number
int count = ((lastnum - firstnum) /
number + 1 );
// return the total sum
return ((lastnum + firstnum) * count) / 2 ;
}
// Driver code
public static void main (String[] args)
{
int n = 3 , number = 7 ;
System.out.println(totalSumDivisibleByNum(n, number));
}
} // This code is contributed by Ajit. |
# Efficient Python3 program to # find the sum divisible numbers. # find the Sum of having n digit # and divisible by the number def totalSumDivisibleByNum(digit, number):
# compute the first and last term
firstnum = pow ( 10 , digit - 1 )
lastnum = pow ( 10 , digit)
# first number which is divisible
# by given number
firstnum = (firstnum - firstnum % number) + number
# last number which is divisible
# by given number
lastnum = (lastnum - lastnum % number)
# total divisible number
count = ((lastnum - firstnum) / number + 1 )
# return the total sum
return int (((lastnum + firstnum) * count) / 2 )
# Driver code digit = 3 ; num = 7
print (totalSumDivisibleByNum(digit, num))
# This code is contributed by Smitha Dinesh Semwal |
// Efficient Java program to find the sum // divisible numbers. using System;
class GFG {
// find the Sum of having n digit and
// divisible by the number
static int totalSumDivisibleByNum( int digit,
int number)
{
// compute the first and last term
int firstnum = ( int )Math.Pow(10, digit - 1);
int lastnum = ( int )Math.Pow(10, digit);
// first number which is divisible
// by given number
firstnum = (firstnum - firstnum % number)
+ number;
// last number which is divisible
// by given number
lastnum = (lastnum - lastnum % number);
// total divisible number
int count = ((lastnum - firstnum) /
number + 1);
// return the total sum
return ((lastnum + firstnum) * count) / 2;
}
// Driver code
public static void Main ()
{
int n = 3, number = 7;
Console.WriteLine(totalSumDivisibleByNum(n, number));
}
} // This code is contributed by vt_m. |
<?php // Efficient PHP program to find // the sum divisible numbers. // find the Sum of having n digit and // divisible by the number function totalSumDivisibleByNum( $digit ,
$number )
{ // compute the first and last term
$firstnum = pow(10, $digit - 1);
$lastnum = pow(10, $digit );
// first number which is divisible
// by given number
$firstnum = ( $firstnum - $firstnum % $number )
+ $number ;
// last number which is divisible
// by given number
$lastnum = ( $lastnum - $lastnum % $number );
// total divisible number
$count = (( $lastnum - $firstnum ) /
$number + 1);
// return the total sum
return (( $lastnum + $firstnum ) *
$count ) / 2;
} // Driver Code
$n = 3; $number = 7;
echo totalSumDivisibleByNum( $n , $number );
// This code is contributed by anuj_67. ?> |
<script> // Efficient Javascript program to find // the sum divisible numbers. // Find the Sum of having n digit and // divisible by the number function totalSumDivisibleByNum(digit, number)
{ // Compute the first and last term
let firstnum = Math.pow(10, digit - 1);
let lastnum = Math.pow(10, digit);
// First number which is divisible
// by given number
firstnum = (firstnum - firstnum %
number) + number;
// Last number which is divisible
// by given number
lastnum = (lastnum - lastnum % number);
// Total divisible number
let count = ((lastnum - firstnum) /
number + 1);
// Return the total sum
return ((lastnum + firstnum) * count) / 2;
} // Driver Code let n = 3, number = 7; document.write(totalSumDivisibleByNum(n, number)); // This code is contributed by divyesh072019 </script> |
70336
Time Complexity: O(1)
Auxiliary Space: O(1)