# Smallest and Largest sum of two n-digit numbers

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

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

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++ 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 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 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# 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)); ` `    ``} ` `} `

 ` `

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

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