Check if a number is multiple of 5 without using / and % operators

Difficulty Level : Basic
Last Updated : 06 Mar, 2020

Given a positive number n, write a function isMultipleof5(int n) that returns true if n is multiple of 5, otherwise false. You are not allowed to use % and / operators.

Method 1 (Repeatedly subtract 5 from n)
Run a loop and subtract 5 from n in the loop while n is greater than 0. After the loop terminates, check whether n is 0. If n becomes 0 then n is multiple of 5, otherwise not.

C++

#include <iostream> 
using namespace std; 
  
// assumes that n is a positive integer  
bool isMultipleof5 (int n)  
{  
    while ( n > 0 )  
        n = n - 5;  
  
    if ( n == 0 )  
        return true;  
  
    return false;  
}  
  
// Driver Code  
int main()  
{  
    int n = 19;  
    if ( isMultipleof5(n) == true )  
        cout << n <<" is multiple of 5";  
    else
        cout << n << " is not a multiple of 5";  
  
    return 0;  
}  
  
// This code is contributed by SHUBHAMSINGH10 

C

#include<stdio.h> 
  
// assumes that n is a positive integer  
bool isMultipleof5 (int n) 
{ 
    while ( n > 0 ) 
        n = n - 5; 
  
    if ( n == 0 ) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 19; 
    if ( isMultipleof5(n) == true ) 
        printf("%d is multiple of 5\n", n); 
    else
        printf("%d is not a multiple of 5\n", n); 
  
    return 0; 
} 

Java

// Java code to check if a number 
// is multiple of 5 without using  
// '/' and '%' operators 
class GFG  
{ 
      
// assumes that n is a positive integer  
static boolean isMultipleof5 (int n) 
{ 
    while (n > 0) 
        n = n - 5; 
  
    if (n == 0) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int n = 19; 
    if (isMultipleof5(n) == true) 
        System.out.printf("%d is multiple of 5\n", n); 
    else
        System.out.printf("%d is not a multiple of 5\n", n); 
} 
} 
  
// This code is contributed by Smitha DInesh Semwal 

Python3

# Python code to check 
# if a number is multiple of 
# 5 without using / and % 
  
# Assumes that n is a positive integer  
def isMultipleof5(n): 
      
    while ( n > 0 ): 
        n = n - 5
  
    if ( n == 0 ): 
        return 1
  
    return 0
      
# Driver Code 
i = 19
if ( isMultipleof5(i) == 1 ): 
    print (i, "is multiple of 5") 
else: 
    print (i, "is not a multiple of 5") 
  
# This code is contributed 
# by Sumit Sudhakar 

C#

// C# code to check if a number 
// is multiple of 5 without using / 
// and % operators 
using System; 
  
class GFG  
{ 
      
// assumes that n is a positive integer 
static bool isMultipleof5 (int n) 
{ 
    while (n > 0) 
        n = n - 5; 
  
    if (n == 0) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
public static void Main() 
{ 
    int n = 19; 
    if (isMultipleof5(n) == true) 
        Console.Write(n + " is multiple of 5\n"); 
    else
        Console.Write(n + " is not a multiple of 5\n"); 
} 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// assumes that n is a positive integer  
function isMultipleof5 ($n) 
{ 
    while ( $n > 0 ) 
        $n = $n - 5; 
  
    if ( $n == 0 ) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
    $n = 19; 
    if ( isMultipleof5($n) == true ) 
        echo("$n is multiple of 5"); 
    else
        echo("$n is not a multiple of 5" ); 
          
// This code is contributed by nitin mittal. 
?> 

Output :

19 is not a multiple of 5

Method 2 (Convert to string and check the last character)
Convert n to a string and check the last character of the string. If the last character is ‘5’ or ‘0’ then n is multiple of 5, otherwise not.

C++

#include <bits/stdc++.h> 
using namespace std; 
  
// Assuming that integre takes 4 bytes, there  
// can be maximum 10 digits in a integer 
# define MAX 11 
  
bool isMultipleof5(int n) 
{ 
    char str[MAX]; 
    int len = strlen(str); 
      
    // Check the last character of string 
    if ( str[len - 1] == '5' ||  
        str[len - 1] == '0' ) 
          
        return true; 
      
    return false; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 19; 
    if ( isMultipleof5(n) == true ) 
        cout << n <<" is multiple of 5" <<endl; 
    else
        cout << n <<" is not multiple of 5" <<endl; 
  
    return 0; 
} 
  
// This code is contributed by SHUBHAMSINGH10 

C

#include<stdio.h> 
#include<stdlib.h> 
#include<string.h> 
  
// Assuming that integre takes 4 bytes, there  
// can be maximum 10 digits in a integer 
# define MAX 11 
  
bool isMultipleof5(int n) 
{ 
    char str[MAX]; 
    int len = strlen(str); 
      
    // Check the last character of string 
    if ( str[len - 1] == '5' ||  
         str[len - 1] == '0' ) 
           
        return true; 
      
    return false; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 19; 
    if ( isMultipleof5(n) == true ) 
        printf("%d is multiple of 5\n", n); 
    else
        printf("%d is not a multiple of 5\n", n); 
  
    return 0; 
} 

Java

// Assuming that integer  
// takes 4 bytes, there  
// can be maximum 10 
// digits in a integer 
class GFG 
{ 
static int MAX = 11; 
  
static boolean isMultipleof5(int n) 
{ 
    char str[] = new char[MAX]; 
    int len = str.length; 
      
    // Check the last  
    // character of string 
    if (str[len - 1] == '5' ||  
        str[len - 1] == '0' ) 
          
        return true; 
      
    return false; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int n = 19; 
    if ( isMultipleof5(n) == true ) 
        System.out.println(n +" is multiple " + 
                                       "of 5"); 
    else
        System.out.println(n +" is not a " + 
                           "multiple of 5"); 
} 
} 
  
// This code is contributed by mits 

Python3

# Assuming that integer 
# takes 4 bytes, there  
# can be maximum 10  
# digits in a integer 
MAX = 11; 
  
def isMultipleof5(n): 
    s = str(n); 
    l = len(s); 
      
    # Check the last  
    # character of string 
    if (s[l - 1] == '5' or
        s[l - 1] == '0'): 
        return True; 
    return False; 
  
# Driver Code 
n = 19; 
if (isMultipleof5(n) == True ): 
    print(n, "is multiple of 5"); 
else: 
    print(n, "is not a multiple of 5"); 
  
# This code is contributed by mits 

C#

using System; 
  
// Assuming that integer 
// takes 4 bytes, there  
// can be maximum 10  
// digits in a integer 
class GFG 
{ 
static int MAX = 11; 
  
static bool isMultipleof5(int n) 
{ 
    char[] str = new char[MAX]; 
    int len = str.Length; 
      
    // Check the last 
    // character of string 
    if (str[len - 1] == '5' ||  
        str[len - 1] == '0' ) 
          
        return true; 
      
    return false; 
} 
  
// Driver Code 
static void Main() 
{ 
    int n = 19; 
    if ( isMultipleof5(n) == true ) 
        Console.WriteLine("{0} is " +  
                          "multiple of 5", n); 
    else
        Console.WriteLine("{0} is not a " + 
                          "multiple of 5", n); 
} 
} 
  
// This code is contributed by mits 

PHP

<?php 
// Assuming that integre  
// takes 4 bytes, there  
// can be maximum 10  
// digits in a integer 
$MAX = 11; 
  
function isMultipleof5($n) 
{ 
    global $MAX; 
    $str = (string)$n; 
    $len = strlen($str); 
      
    // Check the last  
    // character of string 
    if ($str[$len - 1] == '5' ||  
        $str[$len - 1] == '0') 
          
        return true; 
      
    return false; 
} 
  
// Driver Code 
$n = 19; 
if (isMultipleof5($n) == true ) 
    echo "$n is multiple of 5"; 
else
    echo "$n is not a multiple of 5"; 
  
// This code is contributed by mits 
?> 

Output:

19 is not a multiple of 5

Thanks to Baban_Rathore for suggesting this method.

Method 3 (Set last digit as 0 and use floating point trick)
A number n can be a mulpile of 5 in two cases. When last digit of n is 5 or 10. If last bit in binary equivalent of n is set (which can be the case when last digit is 5) then we multiply by 2 using n<<=1 to make sure that if the number is multpile of 5 then we have the last digit as 0. Once we do that, our work is to just check if the last digit is 0 or not, which we can do using float and integer comparison trick.

C++

#include <iostream> 
using namespace std; 
bool isMultipleof5(int n)  
{  
    // If n is a multiple of 5 then we  
    // make sure that last digit of n is 0  
    if ( (n & 1) == 1 )  
        n <<= 1;  
      
    float x = n;  
    x = ( (int)(x * 0.1) ) * 10;  
      
    // If last digit of n is 0 then n  
    // will be equal to (int)x  
    if ( (int)x == n )  
        return true;  
  
    return false;  
}  
  
// Driver Code  
int main()  
{  
    int i = 19;  
    if ( isMultipleof5(i) == true )  
        cout << i <<" is multiple of 5\n";  
    else
        cout << i << " is not a multiple of 5\n";  
  
    getchar();  
    return 0;  
}  
  
// This code is contributed by shubhamsingh10 

C

#include<stdio.h> 
  
bool isMultipleof5(int n) 
{ 
    // If n is a multiple of 5 then we 
    // make sure that last digit of n is 0  
    if ( (n & 1) == 1 ) 
        n <<= 1; 
      
    float x = n; 
    x = ( (int)(x * 0.1) ) * 10; 
      
    // If last digit of n is 0 then n  
    // will be equal to (int)x  
    if ( (int)x == n ) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
int main() 
{ 
    int i = 19; 
    if ( isMultipleof5(i) == true ) 
        printf("%d is multiple of 5\n", i); 
    else
        printf("%d is not a multiple of 5\n", i); 
  
    getchar(); 
    return 0; 
} 

Java

// Java code to check if 
// a number is multiple of 
// 5 without using / and % 
class GFG 
{ 
static boolean isMultipleof5(int n) 
{ 
    // If n is a multiple of 5  
    // then we make sure that 
    // last digit of n is 0  
    if ((n & 1) == 1) 
        n <<= 1; 
      
    float x = n; 
    x = ((int)(x * 0.1)) * 10; 
      
    // If last digit of n is  
    // 0 then n will be equal 
    // to (int)x  
    if ((int)x == n) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int i = 19; 
    if (isMultipleof5(i) == true) 
        System.out.println(i + "is multiple of 5"); 
    else
        System.out.println(i + " is not a " +  
                            "multiple of 5"); 
} 
} 
  
// This code is contributed  
// by mits 

Python3

# Python code to check 
# if a number is multiple of 
# 5 without using / and % 
  
def isMultipleof5(n): 
      
    # If n is a multiple of 5 then we 
    # make sure that last digit of n is 0  
    if ( (n & 1) == 1 ): 
        n <<= 1; 
  
    x = n 
    x = ( (int)(x * 0.1) ) * 10
      
    # If last digit of n is 0 
    # then n will be equal to x 
    if ( x == n ): 
        return 1
  
    return 0
      
# Driver Code 
i = 19
if ( isMultipleof5(i) == 1 ): 
    print (i, "is multiple of 5") 
else: 
    print (i, "is not a multiple of 5") 
  
# This code is contributed 
# by Sumit Sudhakar 

C#

// C# code to check if 
// a number is multiple of 
// 5 without using / and % 
class GFG 
{ 
static bool isMultipleof5(int n) 
{ 
    // If n is a multiple of 5  
    // then we make sure that 
    // last digit of n is 0  
    if ((n & 1) == 1) 
        n <<= 1; 
      
    float x = n; 
    x = ((int)(x * 0.1)) * 10; 
      
    // If last digit of n is  
    // 0 then n will be equal 
    // to (int)x  
    if ((int)x == n) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
public static void Main() 
{ 
    int i = 19; 
    if (isMultipleof5(i) == true) 
        System.Console.WriteLine(i + "is multiple of 5"); 
    else
        System.Console.WriteLine(i + " is not a " +  
                                   "multiple of 5"); 
} 
} 
  
// This code is contributed  
// by mits 

PHP

<?php 
function isMultipleof5($n) 
{ 
    // If n is a multiple of 5 
    // then we make sure that  
    // last digit of n is 0  
    if (($n & 1) == 1 ) 
        $n <<= 1; 
      
    $x = $n; 
    $x = ((int)($x * 0.1)) * 10; 
      
    // If last digit of n  
    // is 0 then n will be  
    // equal to (int)x  
    if ( (int)($x) == $n ) 
        return true; 
  
    return false; 
} 
  
// Driver Code 
$i = 19; 
if ( isMultipleof5($i) == true ) 
    echo "$i is multiple of 5\n"; 
else
    echo "$i is not a multiple of 5\n"; 
  
// This code is contributed by mits 
?> 

Output :

19 is not a multiple of 5

Thanks to darkprince for suggesting this method.

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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