# Sum of n digit numbers divisible by a given number

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
There are nine n digit numbers that
are divisible by 7. Numbers are 14+
21 + 28 + 35 + 42 + 49 + .... + 97.

Input : n = 3, number = 7
Output : 70336

Input : n = 3, number = 4
Output : 124200
```

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

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 ` `#include ` `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; ` `} `

## 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 code ` `n ``=` `3``; num ``=` `7` `print``(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

 ` `

Output:

```70336
```

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 ` `#include ` `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; ` `} `

## 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 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 `

## 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

 ` `

Output:

```70336
```

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