Combien_de_numeros_de_mobiles_francais_sont_des_nombres_premiers.ipynb

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   "source": [
    "# Combien de numéros de mobiles français sont des nombres premiers ?\n",
    "\n",
    "A question simple, réponse simple :"
   ]
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Dépendances"
   ]
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   "source": [
    "from sympy import isprime"
   ]
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      "\n",
      "    Test if n is a prime number (True) or not (False). For n < 2^64 the\n",
      "    answer is definitive; larger n values have a small probability of actually\n",
      "    being pseudoprimes.\n",
      "\n",
      "   \n"
     ]
    }
   ],
   "source": [
    "print(isprime.__doc__[:180])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Réponse"
   ]
  },
  {
   "cell_type": "code",
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   "outputs": [],
   "source": [
    "first_number = 6_00_00_00_00\n",
    "last_number = 7_99_99_99_99\n",
    "\n",
    "# test rapide\n",
    "#last_number = first_number + 20\n",
    "\n",
    "all_numbers = range(first_number, last_number + 1)"
   ]
  },
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   "cell_type": "code",
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    "def count_prime_numbers_in_range(some_range):\n",
    "    count = 0\n",
    "    for number in some_range:\n",
    "        if isprime(number):\n",
    "            count += 1\n",
    "    return count"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Conclusion"
   ]
  },
  {
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   "outputs": [
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     "text": [
      "Pour des numéros de téléphones, nombres entre 600000000 et 799999999 (inclus), il y a 9821476 nombres premiers.\n"
     ]
    }
   ],
   "source": [
    "count = count_prime_numbers_in_range(all_numbers)\n",
    "\n",
    "print(f\"Pour des numéros de téléphones, nombres entre {first_number} et {last_number} (inclus), il y a {count} nombres premiers.\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Et donc, on peut calculer la part de nombres premiers parmi les numéros de téléphones mobiles français.\n",
    "De souvenir, c'était environ 5.1%, vérifions :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
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     "name": "stdout",
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     "text": [
      "Pour des numéros de téléphones, nombres entre 600000000 et 799999999 (inclus), il y a 4.910738% nombres premiers.\n"
     ]
    }
   ],
   "source": [
    "total_number = len(all_numbers)\n",
    "\n",
    "print(f\"Pour des numéros de téléphones, nombres entre {first_number} et {last_number} (inclus), il y a {count/total_number:%} nombres premiers.\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Et voilà, c'était simple !"
   ]
  }
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