# Round the given number to nearest multiple of 10

Given a positive integer n, round it to nearest whole number having zero as last digit.
Examples:

```Input : 4722
Output : 4720

Input : 38
Output : 40

Input : 10
Output: 10

```

Approach:
Let’s round down the given number n to the nearest integer which ends with 0 and store this value in a variable a.
a = (n / 10) * 10. So, the round up n (call it b) is b = a + 10.
If n – a > b – n then the answer is b otherwise the answer is a.
Below is the implementation of the above approach:

## C++

 `// CPP program to round the given ` `// integer to a whole number ` `// which ends with zero.` `#include ` `using` `namespace` `std;`   `// function to round the number` `int` `round(``int` `n)` `{` `    ``// Smaller multiple` `    ``int` `a = (n / 10) * 10;` `    `  `    ``// Larger multiple` `    ``int` `b = a + 10;`   `    ``// Return of closest of two` `    ``return` `(n - a > b - n)? b : a;` `}`   `// driver function` `int` `main()` `{` `    ``int` `n = 4722;` `    ``cout << round(n) << endl;` `    ``return` `0;` `}`

## Java

 `// JAVA Code for Round the given number` `// to nearest multiple of 10` `import` `java.util.*;`   `class` `GFG {` `    `  `    ``// function to round the number` `    ``static` `int` `round(``int` `n)` `    ``{` `        ``// Smaller multiple` `        ``int` `a = (n / ``10``) * ``10``;` `         `  `        ``// Larger multiple` `        ``int` `b = a + ``10``;` `     `  `        ``// Return of closest of two` `        ``return` `(n - a > b - n)? b : a;` `    ``}` `    `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `         ``int` `n = ``4722``;` `         ``System.out.println(round(n));` `    ``}` `}`   `// This code is contributed by Arnav Kr. Mandal.`

## Python3

 `# Python3 code to round the given ` `# integer to a whole number ` `# which ends with zero.`   `# function to round the number` `def` `round``( n ):`   `    ``# Smaller multiple` `    ``a ``=` `(n ``/``/` `10``) ``*` `10` `    `  `    ``# Larger multiple` `    ``b ``=` `a ``+` `10` `    `  `    ``# Return of closest of two` `    ``return` `(b ``if` `n ``-` `a > b ``-` `n ``else` `a)`   `# driver code` `n ``=` `4722` `print``(``round``(n))`   `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# Code for Round the given number` `// to nearest multiple of 10` `using` `System;`   `class` `GFG {` `    `  `    ``// function to round the number` `    ``static` `int` `round(``int` `n)` `    ``{` `        ``// Smaller multiple` `        ``int` `a = (n / 10) * 10;` `        `  `        ``// Larger multiple` `        ``int` `b = a + 10;` `    `  `        ``// Return of closest of two` `        ``return` `(n - a > b - n)? b : a;` `    ``}` `    `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{` `        ``int` `n = 4722;` `        ``Console.WriteLine(round(n));` `    ``}` `}`   `// This code is contributed by Vt_m.`

## PHP

 ` ``\$b` `- ``\$n``) ? ``\$b` `: ``\$a``; ` `} `   `// Driver Code ` `\$n` `= 4722; ` `echo` `roundFunation(``\$n``), ``"\n"``; ` `    `  `// This code is contributed by ajit` `?>`

Output

```4720
```

Output:

```4720

```

Another method if n is large:
The above method is good only for Integer or Long MAX value. if the input length is greater then the int or long-range above method does not work.

We can solve the problem using String.

## Java

 `// Java code for above approach` `import` `java.io.*;`   `class` `GFG ` `{` `  `  `  ``// Program to round the number to the ` `  ``// nearest number having one's digit 0` `  ``public` `static` `String round(String s, ``int` `n)` `  ``{` `    ``char``[] c=s.toCharArray();`   `    ``// last character is 0 then return the` `    ``// original string` `    ``if``(c[n-``1``]==``'0'``)` `      ``return` `s;` `    `  `    ``// if last character is ` `    ``// 1 or 2 or 3 or 4 or 5 make it 0` `    ``else` `if``(c[n-``1``] == ``'1'` `|| c[n-``1``] == ``'2'` `|| ` `            ``c[n-``1``] == ``'3'` `|| c[n-``1``] == ``'4'` `|| ` `            ``c[n-``1``] == ``'5'` `)` `    ``{` `      ``c[n-``1``]=``'0'``;` `      ``return` `new` `String(c);` `    ``}` `    ``else` `    ``{` `      ``c[n-``1``]=``'0'``;` `      `  `      ``// process carry ` `      ``for``(``int` `i = n - ``2` `; i >= ``0` `; i--)` `      ``{` `        ``if``(c[i] == ``'9'``)` `          ``c[i]=``'0'``;` `        ``else` `        ``{` `          ``int` `t= c[i] - ``'0'` `+ ``1``;` `          ``c[i]=(``char``)(``48``+t);` `          ``break``;` `        ``}` `      ``} ` `    ``}`   `    ``String s1=``new` `String(c);`   `    ``if``(s1.charAt(``0``) == ``'0'``)` `      ``s1=``"1"``+s1;` `    `  `    ``// return final string` `    ``return` `s1;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `main (String[] args) ` `  ``{`   `    ``String s=``"5748965412485599999874589965999"``;` `    ``int` `n=s.length();` `    `  `    ``// Function Call` `    ``System.out.println(round(s,n));`   `  ``}` `}`

Output

```5748965412485599999874589966000
```

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