Check if a large number is divisible by 5 or not
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++
// C++ program to find if a number is // divisible by 5 or not #include<bits/stdc++.h> using namespace std; // Function to find that number divisible // by 5 or not. The function assumes that // string length is at least one. bool isDivisibleBy5(string str) { int n = str.length(); return ( ((str[n-1]- '0' ) == 0) || ((str[n-1]- '0' ) == 5)); } // Driver code int main() { string str = "76955" ; isDivisibleBy5(str)? cout << "Yes" : cout << "No " ; return 0; } |
Java
// Java program to find if a number is // divisible by 5 or not import java.io.*; class IsDivisible { // Function to find that number divisible // by 5 or not. The function assumes that // string length is at least one. static boolean isDivisibleBy5(String str) { int n = str.length(); return ( ((str.charAt(n- 1 )- '0' ) == 0 ) || ((str.charAt(n- 1 )- '0' ) == 5 )); } // main function public static void main (String[] args) { String str = "76955" ; if (isDivisibleBy5(str)) System.out.println( "Yes" ); else System.out.println( "No" ); } } |
Python3
# Python program to find if a number is # divisible by 5 or not # Function to find that number divisible # by 5 or not. The function assumes that # string length is at least one. def isDivisibleBy5(st) : n = len (st) return ( (st[n - 1 ] = = '0' ) or (st[n - 1 ] = = '5' )) # Driver code st = "76955" if isDivisibleBy5(st) : print ( "Yes" ) else : print ( "No" ) # This code is contributed by Nikita Tiwari. |
C#
// C# program to find if a number is // C# program to find if a number // is divisible by 5 or not. using System; class IsDivisible { // Function to find that number divisible // by 5 or not. The function assumes that // string length is at least one. static bool isDivisibleBy5(String str) { int n = str.Length; return (((str[n - 1] - '0' ) == 0) || ((str[n - 1] - '0' ) == 5)); } // Driver Code public static void Main () { String str = "76955" ; if (isDivisibleBy5(str)) Console.Write( "Yes" ); else Console.Write( "No" ); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP program to find if a number // is divisible by 5 or not // Function to find that number divisible // by 5 or not. The function assumes that // string length is at least one. function isDivisibleBy5( $str ) { $n = strlen ( $str ); return ( (( $str [ $n -1]- '0' ) == 0) || (( $str [ $n -1]- '0' ) == 5)); } // Driver code $str = "76955" ; $x = isDivisibleBy5( $str ) ? "Yes" : "No" ; echo ( $x ); // This code is contributed by Ajit. ?> |
Javascript
<script> // Javascript program to find if a number // is divisible by 5 or not // Function to find that number divisible // by 5 or not. The function assumes that // string length is at least one. function isDivisibleBy5(str) { n = str.length; return (((str[n - 1] - '0' ) == 0) || ((str[n - 1] - '0' ) == 5)); } // Driver Code var str = "76955" ; var x = isDivisibleBy5(str) ? "Yes" : "No" ; document.write(x); // This code is contributed by bunnyram19 </script> |
Output
Yes
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.