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

  • Difficulty Level : Basic
  • Last Updated : 06 Apr, 2021

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.
?>
Javascript



<script>
 
// JavaScript program for the above approach
 
// 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
    let n = 19;
    if (isMultipleof5(n) == true)
        document.write(n + " is multiple of 5\n");
    else
        document.write(n + " is not a multiple of 5\n");
 
</script>
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
?>
Javascript



<script>
 
// Assuming that integer
// takes 4 bytes, there
// can be maximum 10
// digits in a integer
MAX = 11;
 
function isMultipleof5(n)
{
    str = Array(n).fill('');
    var len = str.length;
     
    // Check the last
    // character of string
    if (str[len - 1] == '5' ||
        str[len - 1] == '0' ) 
        return true;
     
    return false;
}
 
// Driver Code
var n = 19;
if ( isMultipleof5(n) == true )
    document.write(n +" is multiple " +
                                   "of 5");
else
    document.write(n +" is not a " +
                       "multiple of 5");
                        
// This code is contributed by Amit Katiyar
</script>
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 multiple 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 multiple 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
?>
Javascript



<script>
// JavaScript code to check if
// a number is multiple of
// 5 without using / and %
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 = ( (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 == true)
        document.write( i + " is multiple of 5\n");
    else
        document.write( i + " is not a multiple of 5\n");
         
      // This code is contributed by simranarora5sos
    </script>
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.
 
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.
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