Dado un número, la tarea es dividir el número por 3. El número de entrada puede ser grande y puede que no sea posible almacenarlo incluso si usamos long long int.
Ejemplos:
Input : n = 769452 Output : Yes Input : n = 123456758933312 Output : No Input : n = 3635883959606670431112222 Output : Yes
Dado que el número de entrada puede ser muy grande, no podemos usar n % 3 para verificar si un número es divisible por 3 o no, especialmente en lenguajes como C/C++. La idea se basa en el siguiente hecho.
Un número es divisible por 3 si la suma de sus dígitos es divisible por 3.
Ilustración:
For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
= 9
Since sum is divisible by 3,
answer is Yes.
¿Como funciona esto?
Let us consider 1332, we can write it as 1332 = 1*1000 + 3*100 + 3*10 + 2 The proof is based on below observation: Remainder of 10i divided by 3 is 1 So powers of 10 only result in value 1. Remainder of "1*1000 + 3*100 + 3*10 + 2" divided by 3 can be written as : 1*1 + 3*1 + 3*1 + 2 = 9 The above expression is basically sum of all digits. Since 9 is divisible by 3, answer is yes.
A continuación se muestra la implementación del hecho anterior:
C++
// C++ program to find if a number is divisible by
// 3 or not
#include<bits/stdc++.h>
using namespace std;
// Function to find that number divisible by 3 or not
int check(string str)
{
// Compute sum of digits
int n = str.length();
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 3 or not
class IsDivisible
{
// Function to find that number
// divisible by 3 or not
static boolean check(String str)
{
// Compute sum of digits
int n = str.length();
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 program to find if a number is
# divisible by 3 or not
# Function to find that number
# divisible by 3 or not
def check(num) :
# Compute sum of digits
digitSum = 0
while num > 0 :
rem = num % 10
digitSum = digitSum + rem
num = num / 10
# Check if sum of digits is
# divisible by 3.
return (digitSum % 3 == 0)
# main function
num = 1332
if(check(num)) :
print ("Yes")
else :
print ("No")
# This code is contributed by Nikita Tiwari.
C#
// C# program to find if a number is
// divisible by 3 or not
using System;
class GFG
{
// Function to find that number
// divisible by 3 or not
static bool check(string str)
{
// Compute sum of digits
int n = str.Length;
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.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to find if a
// number is divisible by
// 3 or not
// Function to find that
// number divisible by 3 or not
function check($str)
{
// Compute sum of digits
$n = strlen($str);
$digitSum = 0;
for ($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";
$x = check($str) ? "Yes" : "No ";
echo($x);
// This code is contributed by Ajit.
?>
Javascript
<script>
// Javascript program to find if a
// number is divisible by
// 3 or not
// Function to find that
// number divisible by 3 or not
function check(str)
{
// Compute sum of digits
let n = str.length;
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";
let x = check(str) ? "Yes" : "No ";
document.write(x);
// This code is contributed by _saurabh_jaiswal.
</script>
Yes
Complejidad de tiempo : O (logn), donde n es el número dado.
Espacio auxiliar: O(1), ya que no estamos utilizando ningún espacio adicional.
Método 2: comprobar que el número dado es divisible por 3 o no mediante el operador de división de módulo «%».
Python3
# Python code
# To check whether the given number is divisible by 3 or not
#input
n=769452
# the above input can also be given as n=input() -> taking input from user
# finding given number is divisible by 3 or not
if int(n)%3==0:
print("Yes")
else:
print("No")
# this code is contributed by gangarajula laxmi
C#
using System;
public class GFG{
static public void Main (){
//input
long n = 769452;
// the above input can also be given as n=input() -> taking input from user
// finding given number is divisible by 7 or not
if (n % 3 == 0)
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
// This code is contributed by laxmigangarajula03
Javascript
<script>
// JavaScript code for the above approach
// To check whether the given number is divisible by 3 or not
//input
var n = 769452
// finding given number is divisible by 3 or not
if (n % 3 == 0)
document.write("Yes")
else
document.write("No")
// This code is contributed by Potta Lokesh
</script>
Yes
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA