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Multiply the given number by 2 such that it is divisible by 10

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  • Difficulty Level : Basic
  • Last Updated : 21 Jul, 2022

Given a number, the only operation allowed is to multiply the number by 2. Calculate the minimum number of operations to make the number divisible by 10.
NOTE: If it is not possible to convert then print -1.
Examples: 
 

Input: 10 
Output:
As the given number is itself divisible by 10, 
the answer is 0.
Input:
Output: -1 
As by multiplying with 2, given no. can’t be 
converted into a number that is divisible by 10, 
therefore the answer is -1.
 

 

Approach: Any given number is divisible by 10 only if the last digit of the number is 0. For this problem, extract the last digit of the input number and check it in the following ways : 
1) If the last digit is 0 then it is already divisible by 10 , so the minimum number of steps is 0.
2) If the last digit is 5 then multiplying it by 2 one time will make it divisible by 10, so the minimum number of steps is 1.
3) If the last digit is an even or odd number (apart from 0 and 5) then multiplying it by 2 any number of times will only produce even number so we can never make it divisible by 10. Therefore the number of steps is -1.
 

C++




// C++ code for finding
// number of operations
#include <bits/stdc++.h>
using namespace std;
 
int multiplyBy2(int n)
{
    int rem, value;
 
    // Find the last digit or remainder
    rem = n % 10;
    switch (rem) {
 
    // If the last digit is 0
    case 0:
        value = 0;
        break;
 
    // If the last digit is 5
    case 5:
        value = 1;
        break;
 
    // If last digit is other
    // than 0 and 5.
    default:
        value = -1;
    }
 
    return value;
}
 
// Driver code
int main()
{
 
    int n = 28;
    cout << multiplyBy2(n) << endl;
 
    n = 255;
    cout << multiplyBy2(n) << endl;
 
  return 0;
}

Java




// JAVA code for finding
// number of operations
import java.io.*;
 
class GFG
{
    static int multiplyBy2(int n)
{
    int rem, value;
 
    // Find the last digit
    // or remainder
    rem = n % 10;
    switch (rem)
    {
 
    // If the last digit is 0
    case 0:
        value = 0;
        break;
 
    // If the last digit is 5
    case 5:
        value = 1;
        break;
 
    // If last digit is other
    // than 0 and 5.
    default:
        value = -1;
    }
 
    return value;
}
 
// Driver code
public static void main (String[] args)
{
    int n = 28;
    System.out.println(multiplyBy2(n));
 
    n = 255;
    System.out.println(multiplyBy2(n));
}
}
 
// This code is contributed
// by shiv_bhakt.

Python3




# Python3 code for finding number
# of operations
def dig(argu):
    switcher = {
        0: 0,
        5: 1,
    }
    return switcher.get(argu, -1)
     
def multiplyBy2(n):
 
    # Find the last digit or remainder
    rem = n % 10;
    return dig(rem);
 
# Driver code
n = 28;
print(multiplyBy2(n));
 
n = 255;
print(multiplyBy2(n));
         
# This code is contributed by mits

C#




// C# code for finding
// number of operations
using System;
 
class GFG
{
    static int multiplyBy2(int n)
    {
    int rem, value;
 
    // Find the last digit
    // or remainder
    rem = n % 10;
    switch (rem)
    {
 
    // If the last
    // digit is 0
    case 0:
        value = 0;
        break;
 
    // If the last
    // digit is 5
    case 5:
        value = 1;
        break;
 
    // If last digit is
    // other than 0 and 5.
    default:
        value = -1;
        break;
    }
 
    return value;
}
 
// Driver code
public static void Main ()
{
    int n = 28;
    Console.WriteLine(multiplyBy2(n));
 
    n = 255;
    Console.WriteLine(multiplyBy2(n));
}
}
 
// This code is contributed
// by shiv_bhakt.

PHP




<?php
// PHP  code for finding
// number of operations
 
function  multiplyBy2($n)
{
     $rem;
     $value;
 
    // Find the last digit or remainder
    $rem = $n % 10;
    switch ($rem) {
 
    // If the last digit is 0
    case 0:
        $value = 0;
        break;
 
    // If the last digit is 5
    case 5:
        $value = 1;
        break;
 
    // If last digit is other
    // than 0 and 5.
    default:
        $value = -1;
    }
 
    return $value;
}
 
// Driver code
    $n = 28;
    echo  multiplyBy2($n),"\n";
 
    $n = 255;
        echo  multiplyBy2($n),"\n";
         
// This code is contributed by aj_36
?>

Javascript




<script>
 
// Javascript code for finding
// number of operations
 
    function multiplyBy2(n)
    {
        var rem, value;
 
        // Find the last digit
        // or remainder
        rem = n % 10;
        switch (rem) {
 
        // If the last digit is 0
        case 0:
            value = 0;
            break;
 
        // If the last digit is 5
        case 5:
            value = 1;
            break;
 
        // If last digit is other
        // than 0 and 5.
        default:
            value = -1;
        }
 
        return value;
    }
 
    // Driver code
     
        var n = 28;
        document.write(multiplyBy2(n)+"<br/>");
 
        n = 255;
        document.write(multiplyBy2(n));
 
// This code is contributed by todaysgaurav
 
</script>

Output: 

-1
1

 

Method: Using nested if-else

C++




// C++ code to multiply given number
// by 2 such that it is divisible by 10
// using nested if else
#include <iostream>
using namespace std;
 
int main() {
int n = 10;
   
  // checking if the number is divisible
// by 10 or not if divisible print 0
if (n % 10 == 0)
    cout << 0;
   
  // if not divisible by 10 then multiply
// it by 2 and check again
else{
  n = n*2;
    if (n % 10 == 0)
        cout<<0;
    else
        cout<<-1;
}
 
    return 0;
}
 
  // This code is contributed by akashish__

Java




// Java code to multiply given number
// by 2 such that it is divisible by 10
// using nested if else
import java.io.*;
 
class GFG {
  public static void main (String[] args) {
    int n = 10;
 
    // checking if the number is divisible
    // by 10 or not if divisible print 0
    if (n % 10 == 0)
      System.out.println("0");
 
    // if not divisible by 10 then multiply
    // it by 2 and check again
    else{
      n = n*2;
      if (n % 10 == 0)
        System.out.println("0");
      else
        System.out.println("-1");
    }
 
  }
}
 
// This code is contributed by laxmigangarajula03.

Python3




# Python code to multiply given number
# by 2 such that it is divisible by 10
# using nested if else
 
 
n = 10
# checking if the number is divisible
# by 10 or not if divisible print 0
if n % 10 == 0:
    print(0)
# if not divisible by 10 then multiply
# it by 2 and check again
else:
    n = n*2
    if n % 10 == 0:
        print(0)
    else:
        print(-1)
 
  # this code is contributed by gangarajula laxmi

C#




// C# code to multiply given number
// by 2 such that it is divisible by 10
// using nested if else
using System;
 
public class GFG{
 
  static public void Main (){
 
    // Code
    int n = 10;
 
    // checking if the number is divisible
    // by 10 or not if divisible print 0
    if (n % 10 == 0)
    { Console.WriteLine("0"); }
 
    // if not divisible by 10 then multiply
    // it by 2 and check again
    else{
      n = n*2;
      if (n % 10 == 0)
      { Console.WriteLine("0"); }
      else
      { Console.WriteLine("-1"); }
    }
  }
}
// This code is contributed by akashish__

Javascript




<script>
      // JavaScript code for the above approach
      let n = 10;
 
      // checking if the number is divisible
      // by 10 or not if divisible print 0
      if (n % 10 == 0)
          document.write(0);
 
      // if not divisible by 10 then multiply
      // it by 2 and check again
      else {
          n = n * 2;
          if (n % 10 == 0)
              document.write(0);
          else
              document.write(-1);
      }
 
  // This code is contributed by Potta Lokesh
  </script>

Output

0

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