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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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## 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 ` `?> ` |

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**Output:**

2