Dado un número, la tarea es verificar si un número es divisible por 6 o no. El número de entrada puede ser grande y puede que no sea posible almacenarlo incluso si usamos long long int.
Ejemplos:
Input : n = 2112 Output: Yes Input : n = 1124 Output : No Input : n = 363588395960667043875487 Output : No
Python3
# Python code
# To check whether the given number is divisible by 6 or not
#input
n=363588395960667043875487
# the above input can also be given as n=input() -> taking input from user
# finding given number is divisible by 6 or not
if int(n)%6==0:
print("Yes")
else:
print("No")
Javascript
<script>
// JavaScript code for the above approach
// To check whether the given number is divisible by 6 or not
//input
var n = 363588395960667043875487
// finding given number is divisible by 6 or not
if (n % 6 == 0)
document.write("Yes")
else
document.write("No")
// This code is contributed by Potta Lokesh
</script>
Producción
No
Dado que el número de entrada puede ser muy grande, no podemos usar n % 6 para verificar si un número es divisible por 6 o no, especialmente en lenguajes como C/C++. La idea se basa en el siguiente hecho.
A number is divisible by 6 it's divisible by 2 and 3. a) A number is divisible by 2 if its last digit is divisible by 2. b) A number is divisible by 3 if sum of digits is divisible by 3.
A continuación se muestra la implementación basada en los pasos anteriores.
C++
// C++ program to find if a number is divisible by
// 6 or not
#include<bits/stdc++.h>
using namespace std;
// Function to find that number divisible by 6 or not
bool check(string str)
{
int n = str.length();
// Return false if number is not divisible by 2.
if ((str[n-1]-'0')%2 != 0)
return false;
// If we reach here, number is divisible by 2.
// Now check for 3.
// Compute sum of digits
int digitSum = 0;
for (int i=0; i<n; i++)
digitSum += (str[i]-'0');
// Check if sum of digits is divisible by 3
return (digitSum % 3 == 0);
}
// Driver code
int main()
{
string str = "1332";
check(str)? cout << "Yes" : cout << "No ";
return 0;
}
Java
// Java program to find if a number is
// divisible by 6 or not
class IsDivisible
{
// Function to find that number divisible by 6 or not
static boolean check(String str)
{
int n = str.length();
// Return false if number is not divisible by 2.
if ((str.charAt(n-1) -'0')%2 != 0)
return false;
// If we reach here, number is divisible by 2.
// Now check for 3.
// Compute sum of digits
int digitSum = 0;
for (int i=0; i<n; i++)
digitSum += (str.charAt(i)-'0');
// Check if sum of digits is divisible by 3
return (digitSum % 3 == 0);
}
// main function
public static void main (String[] args)
{
String str = "1332";
if(check(str))
System.out.println("Yes");
else
System.out.println("No");
}
}
Python3
# Python 3 program to find
# if a number is divisible
# by 6 or not
# Function to find that number
# is divisible by 6 or not
def check(st) :
n = len(st)
# Return false if number
# is not divisible by 2.
if (((int)(st[n-1])%2) != 0) :
return False
# If we reach here, number
# is divisible by 2. Now
# check for 3.
# Compute sum of digits
digitSum = 0
for i in range(0, n) :
digitSum = digitSum + (int)(st[i])
# Check if sum of digits
# is divisible by 3
return (digitSum % 3 == 0)
# Driver code
st = "1332"
if(check(st)) :
print("Yes")
else :
print("No ")
# This article is contributed by Nikita Tiwari.
C#
// C# program to find if a number is
// divisible by 6 or not
using System;
class GFG {
// Function to find that number
// divisible by 6 or not
static bool check(String str)
{
int n = str.Length;
// Return false if number is
// not divisible by 2.
if ((str[n-1] -'0') % 2 != 0)
return false;
// If we reach here, number is
// divisible by 2.
// Now check for 3.
// Compute sum of digits
int digitSum = 0;
for (int i = 0; i < n; i++)
digitSum += (str[i] - '0');
// Check if sum of digits is
// divisible by 3
return (digitSum % 3 == 0);
}
// main function
public static void Main ()
{
String str = "1332";
if(check(str))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by parashar.
PHP
<?php
// PHP program to find if a
// number is divisible by
// 6 or not
// Function to find that number
// divisible by 6 or not
function check($str)
{
$n = strlen($str);
// Return false if number is
// not divisible by 2.
if (($str[$n - 1] - '0') % 2 != 0)
return false;
// If we reach here, number
// is divisible by 2.
// Now check for 3.
// Compute sum of digits
$digitSum = 0;
for ($i = 0; $i < $n; $i++)
$digitSum += ($str[$i] - '0');
// Check if sum of digits
// is divisible by 3
return ($digitSum % 3 == 0);
}
// Driver code
$str = "1332";
if(check($str))
echo "Yes" ;
else
echo " No ";
return 0;
// This code is contributed by nitin mittal.
?>
Javascript
<script>
// JavaScript program for the above approach
// Function to find that number
// divisible by 6 or not
function check(str)
{
let n = str.length;
// Return false if number is
// not divisible by 2.
if ((str[n-1] -'0') % 2 != 0)
return false;
// If we reach here, number is
// divisible by 2.
// Now check for 3.
// Compute sum of digits
let digitSum = 0;
for (let i = 0; i < n; i++)
digitSum += (str[i] - '0');
// Check if sum of digits is
// divisible by 3
return (digitSum % 3 == 0);
}
// Driver Code
let str = "1332";
if(check(str))
document.write("Yes");
else
document.write("No");
// This code is contributed by splevel62.
</script>
Producción
Yes
Complejidad de tiempo: O(logN) donde N es el número dado
Espacio auxiliar: O(1) ya que no se utiliza espacio adicional
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA