Check if a large number is divisible by 5 or not
Last Updated :
13 Sep, 2023
Given a number, the task is to check if number is divisible by 5. The input number may be large and it may not be possible to store even if we use long long int.
Examples:
Input : n = 56945255
Output : Yes
Input : n = 1234567589333150
Output : Yes
Input : n = 3635883959606670431112222
Output : No
Since input number may be very large, we cannot use n % 5 to check if a number is divisible by 5 or not, especially in languages like C/C++. The idea is based on following fact.
A number is divisible by 5 if its
digits last digit will be 0 or 5 .
Illustration:
For example, let us consider 769555
Number formed by last digit is = 5
Since 5 is divisible by 5 , answer is YES.
How does this work?
Let us consider 5335, we can write it as
5335 = 5*1000 + 3*100 + 3*10 + 5
The proof is based on below observation:
Remainder of 10i divided by 5 is 0 if i greater
than or equal to one. Note than 10, 100, 1000,
... etc lead to remainder 0 when divided by 5.
So remainder of " 5*1000 + 3*100 +
3*10 + 5" divided by 5 is equivalent to remainder
of following :
0 + 0 + 0 + 5 = 5
Since 5 is divisible by 5, answer is yes.
Below is the implementation of above idea.
C++
#include<bits/stdc++.h>
using namespace std;
bool isDivisibleBy5(string str)
{
int n = str.length();
return ( ((str[n-1]-'0') == 0) ||
((str[n-1]-'0') == 5));
}
int main()
{
string str = "76955";
isDivisibleBy5(str)? cout << "Yes" :
cout << "No ";
return 0;
}
|
Java
import java.io.*;
class IsDivisible
{
static boolean isDivisibleBy5(String str)
{
int n = str.length();
return ( ((str.charAt(n-1)-'0') == 0) ||
((str.charAt(n-1)-'0') == 5));
}
public static void main (String[] args)
{
String str = "76955";
if(isDivisibleBy5(str))
System.out.println("Yes");
else
System.out.println("No");
}
}
|
Python3
def isDivisibleBy5(st) :
n = len(st)
return ( (st[n-1] == '0') or
(st[n-1] == '5'))
st = "76955"
if isDivisibleBy5(st) :
print ("Yes")
else :
print ("No")
|
C#
using System;
class IsDivisible
{
static bool isDivisibleBy5(String str)
{
int n = str.Length;
return (((str[n - 1] - '0') == 0) ||
((str[n - 1] - '0') == 5));
}
public static void Main ()
{
String str = "76955";
if(isDivisibleBy5(str))
Console.Write("Yes");
else
Console.Write("No");
}
}
|
PHP
<?php
function isDivisibleBy5($str)
{
$n = strlen($str);
return ( (($str[$n-1]-'0') == 0) ||
(($str[$n-1]-'0') == 5));
}
$str = "76955";
$x = isDivisibleBy5($str) ? "Yes" : "No";
echo($x);
?>
|
Javascript
<script>
function isDivisibleBy5(str)
{
n = str.length;
return (((str[n - 1] - '0') == 0) ||
((str[n - 1] - '0') == 5));
}
var str = "76955";
var x = isDivisibleBy5(str) ? "Yes" : "No";
document.write(x);
</script>
|
Time Complexity: O(1), as we are not using any loops for traversing.
Auxiliary Space: O(1), as we are not using any extra space.
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