# Smallest and Largest sum of two n-digit numbers

• Last Updated : 26 Mar, 2021

Given an integer N ≥ 1, the task is to find the smallest and the largest sum of two N digit numbers.
Examples:

Input: N = 1
Output:
Largest = 18
Smallest = 0
Largest 1-digit number is 9 and 9 + 9 = 18
Smallest 1-digit number is 0 and 0 + 0 = 0
Input: N = 2
Output:
Largest = 198
Smallest = 20

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Approach:

• For largest: The answer will be 2 * (10N – 1) because the series of sum of two n digit numbers will go on like 2 * 9, 2 * 99, 2 * 999, …
• For smallest:
• If N = 1 then the answer will be 0.
• If N > 1 then the answer will be 2 * (10N – 1) because the series of sum of two n digit numbers will go on like 0, 20, 200, 2000, …

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the smallest sum``// of 2 n-digit numbers``int` `smallestSum(``int` `n)``{``    ``if` `(n == 1)``        ``return` `0;``    ``return` `(2 * ``pow``(10, n - 1));``}` `// Function to return the largest sum``// of 2 n-digit numbers``int` `largestSum(``int` `n)``{``    ``return` `(2 * (``pow``(10, n) - 1));``}` `// Driver code``int` `main()``{``    ``int` `n = 4;``    ``cout << ``"Largest = "` `<< largestSum(n) << endl;``    ``cout << ``"Smallest = "` `<< smallestSum(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG {` `    ``// Function to return the smallest sum``    ``// of 2 n-digit numbers``    ``static` `int` `smallestSum(``int` `n)``    ``{``        ``if` `(n == ``1``)``            ``return` `0``;``        ``return` `(``2` `* (``int``)Math.pow(``10``, n - ``1``));``    ``}` `    ``// Function to return the largest sum``    ``// of 2 n-digit numbers``    ``static` `int` `largestSum(``int` `n)``    ``{``        ``return` `(``2` `* ((``int``)Math.pow(``10``, n) - ``1``));``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n = ``4``;``        ``System.out.println(``"Largest = "` `+ largestSum(n));``        ``System.out.print(``"Smallest = "` `+ smallestSum(n));``    ``}``}`

## Python3

 `# Python3 implementation of the approach` `# Function to return the smallest sum``# of 2 n-digit numbers``def` `smallestSum(n):`` ` `    ``if` `(n ``=``=` `1``):``        ``return` `0``    ``return` `(``2` `*` `pow``(``10``, n ``-` `1``))` `# Function to return the largest sum``# of 2 n-digit numbers``def` `largestSum(n):``    ``return` `(``2` `*` `(``pow``(``10``, n) ``-` `1``))` `# Driver code``n ``=` `4``print``(``"Largest = "``, largestSum(n))``print``(``"Smallest = "``, smallestSum(n))`

## C#

 `// C# implementation of the approach``using` `System;``class` `GFG {` `    ``// Function to return the smallest sum``    ``// of 2 n-digit numbers``    ``static` `int` `smallestSum(``int` `n)``    ``{``        ``if` `(n == 1)``            ``return` `0;``        ``return` `(2 * (``int``)Math.Pow(10, n - 1));``    ``}` `    ``// Function to return the largest sum``    ``// of 2 n-digit numbers``    ``static` `int` `largestSum(``int` `n)``    ``{``        ``return` `(2 * ((``int``)Math.Pow(10, n) - 1));``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 4;``        ``Console.WriteLine(``"Largest = "` `+ largestSum(n));``        ``Console.Write(``"Smallest = "` `+ smallestSum(n));``    ``}``}`

## PHP

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## Javascript

 ``
Output:
```Largest = 19998
Smallest = 2000```

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