Comment Déterminer Rapidement si un Nombre est un Emirp : Guide Complet et Exemples

#include <iostream>
#include <cmath>

bool isPrime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0) return false;
    return true;
}

int reverseNumber(int n) {
    int reversed = 0;
    while (n > 0) {
        reversed = reversed * 10 + n % 10;
        n /= 10;
    }
    return reversed;
}

bool isEmirp(int n) {
    return isPrime(n) && isPrime(reverseNumber(n)) && (n != reverseNumber(n));
}

int main() {
    int num = 13;
    if (isEmirp(num))
        std::cout << "Oui" << std::endl;
    else
        std::cout << "Non" << std::endl;
    return 0;
}

class Emirp {
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i <= Math.sqrt(n); i++)
            if (n % i == 0) return false;
        return true;
    }

    public static int reverseNumber(int n) {
        int reversed = 0;
        while (n > 0) {
            reversed = reversed * 10 + n % 10;
            n /= 10;
        }
        return reversed;
    }

    public static boolean isEmirp(int n) {
        return isPrime(n) && isPrime(reverseNumber(n)) && (n != reverseNumber(n));
    }

    public static void main(String[] args) {
        int num = 13;
        if (isEmirp(num))
            System.out.println("Oui");
        else
            System.out.println("Non");
    }
}

def is_prime(n):
    if n <= 1:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False
    return True

def reverse_number(n):
    return int(str(n)[::-1])

def is_emirp(n):
    return is_prime(n) and is_prime(reverse_number(n)) and (n != reverse_number(n))

num = 13
if is_emirp(num):
    print("Oui")
else:
    print("Non")

using System;

public class Emirp {
    public static bool IsPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i <= Math.Sqrt(n); i++)
            if (n % i == 0) return false;
        return true;
    }

    public static int ReverseNumber(int n) {
        int reversed = 0;
        while (n > 0) {
            reversed = reversed * 10 + n % 10;
            n /= 10;
        }
        return reversed;
    }

    public static bool IsEmirp(int n) {
        return IsPrime(n) && IsPrime(ReverseNumber(n)) && (n != ReverseNumber(n));
    }

    public static void Main(string[] args) {
        int num = 13;
        if (IsEmirp(num))
            Console.WriteLine("Oui");
        else
            Console.WriteLine("Non");
    }
}

function isPrime(n) {
    if (n <= 1) return false;
    for (let i = 2; i <= Math.sqrt(n); i++)
        if (n % i === 0) return false;
    return true;
}

function reverseNumber(n) {
    return parseInt(n.toString().split('').reverse().join(''));
}

function isEmirp(n) {
    return isPrime(n) && isPrime(reverseNumber(n)) && (n != reverseNumber(n));
}

let num = 13;
if (isEmirp(num))
    console.log("Oui");
else
    console.log("Non");

<?php
function isPrime($n) {
    if ($n <= 1) return false;
    for ($i = 2; $i <= sqrt($n); $i++)
        if ($n % $i == 0) return false;
    return true;
}

function reverseNumber($n) {
    return (int)strrev((string)$n);
}

function isEmirp($n) {
    return isPrime($n) && isPrime(reverseNumber($n)) && ($n != reverseNumber($n));
}

$num = 13;
if (isEmirp($num))
    echo "Oui";
else
    echo "Non";
?>