# n-th number whose sum of digits is ten

Given an integer value n, find out the n-th positive integer whose sum is 10.

Examples:

```Input: n = 2
Output: 28
The first number with sum of digits as
10 is 19. Second number is 28.
```
```Input: 15
Output: 154
```

Method 1 (Simple):
We traverse through all numbers. For every number, we find the sum of digits. We stop when we find the n-th number with the sum of digits as 10.

## C++

 `// Simple CPP program to find n-th number` `// with sum of digits as 10.` `#include ` `using` `namespace` `std;`   `int` `findNth(``int` `n)` `{` `    ``int` `count = 0;`   `    ``for` `(``int` `curr = 1;; curr++) {`   `        ``// Find sum of digits in current no.` `        ``int` `sum = 0;` `        ``for` `(``int` `x = curr; x > 0; x = x / 10)` `            ``sum = sum + x % 10;`   `        ``// If sum is 10, we increment count` `        ``if` `(sum == 10)` `            ``count++;`   `        ``// If count becomes n, we return current` `        ``// number.` `        ``if` `(count == n)` `            ``return` `curr;` `    ``}` `    ``return` `-1;` `}`   `int` `main()` `{` `    ``printf``(``"%d\n"``, findNth(5));` `    ``return` `0;` `}`

## Java

 `// Java program to find n-th number` `// with sum of digits as 10.`   `import` `java.util.*;` `import` `java.lang.*;`   `public` `class` `GFG {` `    ``public` `static` `int` `findNth(``int` `n)` `    ``{` `        ``int` `count = ``0``;` `        ``for` `(``int` `curr = ``1``;; curr++) {`   `            ``// Find sum of digits in current no.` `            ``int` `sum = ``0``;` `            ``for` `(``int` `x = curr; x > ``0``; x = x / ``10``)` `                ``sum = sum + x % ``10``;`   `            ``// If sum is 10, we increment count` `            ``if` `(sum == ``10``)` `                ``count++;`   `            ``// If count becomes n, we return current` `            ``// number.` `            ``if` `(count == n)` `                ``return` `curr;` `        ``}` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``System.out.print(findNth(``5``));` `    ``}` `}`   `// Contributed by _omg`

## Python3

 `# Python3 program to find n-th number` `# with sum of digits as 10.` `import` `itertools`   `# function to find required number` `def` `findNth(n):`   `    ``count ``=` `0`   `    ``for` `curr ``in` `itertools.count():` `        ``# Find sum of digits in current no.` `        ``sum` `=` `0` `        ``x ``=` `curr` `        ``while``(x):` `            ``sum` `=` `sum` `+` `x ``%` `10` `            ``x ``=` `x ``/``/` `10`   `        ``# If sum is 10, we increment count` `        ``if` `(``sum` `=``=` `10``):` `            ``count ``=` `count ``+` `1`   `        ``# If count becomes n, we return current` `        ``# number.` `        ``if` `(count ``=``=` `n):` `            ``return` `curr`   `    ``return` `-``1`   `# Driver program` `if` `__name__``=``=``'__main__'``:` `    ``print``(findNth(``5``))`   `# This code is contributed by` `# Sanjit_Prasad`

## C#

 `// C# program to find n-th number` `// with sum of digits as 10.` `using` `System;`   `class` `GFG {` `    ``public` `static` `int` `findNth(``int` `n)` `    ``{` `        ``int` `count = 0;` `        ``for` `(``int` `curr = 1;; curr++) {`   `            ``// Find sum of digits in current no.` `            ``int` `sum = 0;` `            ``for` `(``int` `x = curr; x > 0; x = x / 10)` `                ``sum = sum + x % 10;`   `            ``// If sum is 10, we increment count` `            ``if` `(sum == 10)` `                ``count++;`   `            ``// If count becomes n, we` `            ``// return current number.` `            ``if` `(count == n)` `                ``return` `curr;` `        ``}` `    ``}`   `    ``// Driver Code` `    ``static` `public` `void` `Main()` `    ``{` `        ``Console.WriteLine(findNth(5));` `    ``}` `}`   `// This code is contributed` `// by Sach_Code`

## Javascript

 ``

## PHP

 ` 0; ``\$x` `= ``\$x` `/ 10) ` `            ``\$sum` `= ``\$sum` `+ ``\$x` `% 10; `   `        ``// If sum is 10, we increment ` `        ``// count ` `        ``if` `(``\$sum` `== 10) ` `            ``\$count``++; `   `        ``// If count becomes n, we return ` `        ``// current number. ` `        ``if` `(``\$count` `== ``\$n``) ` `            ``return` `\$curr``; ` `    ``} ` `    ``return` `-1; ` `} `   `// Driver Code` `echo` `findNth(5); `   `// This code is contributed by Sach .` `?>`

Output

```55
```

Time Complexity: O(N*log10(N))
Auxiliary Space: O(1)

### Method 2 (Efficient):

If we take a closer look, we can notice that all multiples of 9 are present in arithmetic progression 19, 28, 37, 46, 55, 64, 73, 82, 91, 100, 109,… However, there are numbers in the above series whose sum of digits is not 10, for example, 100. So instead of checking one by one, we start with 19 and increment by 9.

## C++

 `// Simple CPP program to find n-th number` `// with sum of digits as 10.` `#include ` `using` `namespace` `std;`   `int` `findNth(``int` `n)` `{` `    ``int` `count = 0;`   `    ``for` `(``int` `curr = 19;; curr += 9) {`   `        ``// Find sum of digits in current no.` `        ``int` `sum = 0;` `        ``for` `(``int` `x = curr; x > 0; x = x / 10)` `            ``sum = sum + x % 10;`   `        ``// If sum is 10, we increment count` `        ``if` `(sum == 10)` `            ``count++;`   `        ``// If count becomes n, we return current` `        ``// number.` `        ``if` `(count == n)` `            ``return` `curr;` `    ``}` `    ``return` `-1;` `}`   `int` `main()` `{` `    ``printf``(``"%d\n"``, findNth(5));` `    ``return` `0;` `}`

## Java

 `// Java program to find n-th number` `// with sum of digits as 10.`   `import` `java.util.*;` `import` `java.lang.*;`   `public` `class` `GFG {` `    ``public` `static` `int` `findNth(``int` `n)` `    ``{` `        ``int` `count = ``0``;`   `        ``for` `(``int` `curr = ``19``;; curr += ``9``) {`   `            ``// Find sum of digits in current no.` `            ``int` `sum = ``0``;` `            ``for` `(``int` `x = curr; x > ``0``; x = x / ``10``)` `                ``sum = sum + x % ``10``;`   `            ``// If sum is 10, we increment count` `            ``if` `(sum == ``10``)` `                ``count++;`   `            ``// If count becomes n, we return current` `            ``// number.` `            ``if` `(count == n)` `                ``return` `curr;` `        ``}` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``System.out.print(findNth(``5``));` `    ``}` `}`   `// Contributed by _omg`

## Python3

 `# Python3 program to find n-th ` `# number with sum of digits as 10. ` `def` `findNth(n): ` `    ``count ``=` `0``;` `    `  `    ``curr ``=` `19``;`   `    ``while` `(``True``): `   `        ``# Find sum of digits in` `        ``# current no. ` `        ``sum` `=` `0``;` `        ``x ``=` `curr;` `        ``while` `(x > ``0``):` `            ``sum` `=` `sum` `+` `x ``%` `10``;` `            ``x ``=` `int``(x ``/` `10``);`   `        ``# If sum is 10, we increment` `        ``# count ` `        ``if` `(``sum` `=``=` `10``): ` `            ``count``+``=` `1``; `   `        ``# If count becomes n, we return ` `        ``# current number. ` `        ``if` `(count ``=``=` `n): ` `            ``return` `curr;` `        `  `        ``curr ``+``=` `9``;`   `    ``return` `-``1``; `   `# Driver Code` `print``(findNth(``5``)); `   `# This code is contributed ` `# by mits`

## C#

 `// C# program to find n-th number` `// with sum of digits as 10.` `using` `System;`   `class` `GFG {` `    ``public` `static` `int` `findNth(``int` `n)` `    ``{` `        ``int` `count = 0;`   `        ``for` `(``int` `curr = 19;; curr += 9) {`   `            ``// Find sum of digits in` `            ``// current no.` `            ``int` `sum = 0;` `            ``for` `(``int` `x = curr;` `                 ``x > 0; x = x / 10)` `                ``sum = sum + x % 10;`   `            ``// If sum is 10, we increment` `            ``// count` `            ``if` `(sum == 10)` `                ``count++;`   `            ``// If count becomes n, we return` `            ``// current number.` `            ``if` `(count == n)` `                ``return` `curr;` `        ``}` `    ``}`   `    ``// Driver Code` `    ``static` `public` `void` `Main()` `    ``{` `        ``Console.WriteLine(findNth(5));` `    ``}` `}`   `// This code is contributed` `// by Sach_Code`

## Javascript

 ``

## PHP

 ` 0;` `             ``\$x` `= (int)``\$x` `/ 10) ` `            ``\$sum` `= ``\$sum` `+ ``\$x` `% 10; `   `        ``// If sum is 10, we increment` `        ``// count ` `        ``if` `(``\$sum` `== 10) ` `            ``\$count``++; `   `        ``// If count becomes n, we return ` `        ``// current number. ` `        ``if` `(``\$count` `== ``\$n``) ` `            ``return` `\$curr``; ` `    ``} ` `    ``return` `-1; ` `} `   `// Driver Code` `echo` `findNth(5); `   `// This code is contributed ` `// by Sach_Code` `?>`

Output

```55
```

Time Complexity: O(N*log10(N))
Auxiliary Space: O(1)

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