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.

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 <bits/stdc++.h>` `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

`<?php` `// See if we can buy less than 10 items` `// Using 10 Rs coins and one r Rs coin` `function` `minItems(` `$k` `, ` `$r` `)` `{` ` ` `for` `(` `$i` `= 1; ` `$i` `< 10; ` `$i` `++)` ` ` `if` `((` `$i` `* ` `$k` `- ` `$r` `) % 10 == 0 ||` ` ` `(` `$i` `* ` `$k` `) % 10 == 0)` ` ` `return` `$i` `;` ` ` ` ` `// We can always buy 10 items` ` ` `return` `10;` `}` `// Driver Code` `$k` `= 15;` `$r` `= 2;` `echo` `minItems(` `$k` `, ` `$r` `);` `// This code is contributed by Rajput-Ji` `?>` |

## Javascript

`<script>` `// Javascript program of the above approach` `function` `minItems(k, r)` `{` ` ` `// See if we can buy less than 10 items` `// Using 10 Rs coins and one r Rs coin` `for` `(let 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` ` ` `let k = 15;` ` ` `let r = 2;` ` ` `document.write(minItems(k, r));` `</script>` |

**Output:**

2

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