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

04 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: 0 As the given number is itself divisible by 10, the answer is 0.Input: 1 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++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){     int n = 28;    cout << multiplyBy2(n) << endl;     n = 255;    cout << multiplyBy2(n) << endl;   return 0;}
// JAVA code for finding// number of operationsimport 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 codepublic 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 code for finding number# of operationsdef 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 coden = 28;print(multiplyBy2(n)); n = 255;print(multiplyBy2(n));         # This code is contributed by mits
// C# code for finding// number of operationsusing 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 codepublic 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  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?>
<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>
-1
1
 
C++
Python3
C#
Javascript
// 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 0if (n % 10 == 0)    cout << 0;     // if not divisible by 10 then multiply// it by 2 and check againelse{  n = n*2;    if (n % 10 == 0)        cout<<0;    else        cout<<-1;}     return 0;}   // This code is contributed by akashish__
# 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 0if n % 10 == 0:    print(0)# if not divisible by 10 then multiply# it by 2 and check againelse:    n = n*2    if n % 10 == 0:        print(0)    else:        print(-1)   # this code is contributed by gangarajula laxmi
// C# code to multiply given number// by 2 such that it is divisible by 10// using nested if elseusing 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__
<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>
0
jit_t
Vishal_Khoda
Mithun Kumar
todaysgaurav
laxmigangarajula03
akashish__
lokeshpotta20
school-programming
Mathematical
Mathematical
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