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.
C++
// 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 codeint main(){ int n = 3, num = 7; cout << totalSumDivisibleByNum(n, num) << "\n"; return 0;} |
Java
// 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. |
Python3
# 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 coden = 3; num = 7print(totalSumDivisibleByNum(n, num))# This code is contributed by Smitha Dinesh Semwal |
C#
// 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
<?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?> |
Javascript
<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 Codelet n = 3, num = 7;document.write(totalSumDivisibleByNum(n, num));// This code is contributed by chinmoy1997pal </script> |
Output:
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)
C++
// 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 numberint 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;} |
Java
// 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. |
Python3
# Efficient Python3 program to # find the sum divisible numbers.# find the Sum of having n digit # and divisible by the numberdef 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 codedigit = 3; num = 7print(totalSumDivisibleByNum(digit, num))# This code is contributed by Smitha Dinesh Semwal |
C#
// 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
<?php// Efficient PHP program to find // the sum divisible numbers.// find the Sum of having n digit and// divisible by the numberfunction 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.?> |
Javascript
<script>// Efficient Javascript program to find// the sum divisible numbers.// Find the Sum of having n digit and// divisible by the numberfunction 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 Codelet n = 3, number = 7; document.write(totalSumDivisibleByNum(n, number));// This code is contributed by divyesh072019</script> |
Output:
70336
Time Complexity: O(1)
Auxiliary Space: O(1)
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