# Buy minimum items without change and given coins

You have an unlimited number of 10-rupee coins and exactly one coin of r rupee and you need to buy minimum items each of cost k such that you do not ask for change.

Examples:

Input: k = 15, r = 2
Output: 2
You should buy two cables and pay 2*15=30 rupees. It is obvious that you can pay this sum without any change.

Input: k = 237, r = 7
Output:1
It is enough for you to buy one cable.

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

It is obvious that we can pay for 10 items without any change (by paying the required amount of 10-rupee coins and not using the coin of r rupee). But perhaps you can buy fewer hammers and pay without any change. Note that you should buy at least one item.

## C++

 `#include ` `using` `namespace` `std; ` ` `  `int` `minItems(``int` `k, ``int` `r) ` `{ ` `   ``// See if we can buy less than 10 items ` `   ``// Using 10 Rs coins and one r Rs coin ` `   ``for` `(``int` `i = 1; i < 10; i++)  ` `        ``if` `((i * k - r) % 10 == 0 || ` `            ``(i * k) % 10 == 0)  ` `            ``return` `i; ` ` `  `    ``// We can always buy 10 items ` `    ``return` `10; ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `k = 15; ` `    ``int` `r = 2; ` `    ``cout << minItems(k, r); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `static` `int` `minItems(``int` `k, ``int` `r) ` `{ ` `     `  `// See if we can buy less than 10 items ` `// Using 10 Rs coins and one r Rs coin ` `for` `(``int` `i = ``1``; i < ``10``; i++)  ` `        ``if` `((i * k - r) % ``10` `== ``0` `|| ` `            ``(i * k) % ``10` `== ``0``)  ` `            ``return` `i; ` ` `  `    ``// We can always buy 10 items ` `    ``return` `10``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `k = ``15``; ` `    ``int` `r = ``2``; ` `    ``System.out.println(minItems(k, r)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by SURENDRA_GANGWAR `

## Python3

 `# Python3 implementation of above approach ` ` `  `def` `minItems(k, r) : ` ` `  `    ``# See if we can buy less than 10 items  ` `    ``# Using 10 Rs coins and one r Rs coin  ` `    ``for` `i ``in` `range``(``1``, ``10``) :  ` `            ``if` `((i ``*` `k ``-` `r) ``%` `10` `=``=` `0` `or`  `                ``(i ``*` `k) ``%` `10` `=``=` `0``) : ` `                ``return` `i  ` `     `  `    ``# We can always buy 10 items  ` `    ``return` `10``;  ` `     `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``k, r ``=` `15` `, ``2``;  ` `    ``print``(minItems(k, r)) ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `static` `int` `minItems(``int` `k, ``int` `r) ` `{ ` `     `  `// See if we can buy less than 10 items ` `// Using 10 Rs coins and one r Rs coin ` `for` `(``int` `i = 1; i < 10; i++)  ` `        ``if` `((i * k - r) % 10 == 0 || ` `            ``(i * k) % 10 == 0)  ` `            ``return` `i; ` ` `  `    ``// We can always buy 10 items ` `    ``return` `10; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `k = 15; ` `    ``int` `r = 2; ` `    ``Console.WriteLine(minItems(k, r)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by inder_verma `

## PHP

 ` `

Output:

```2
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