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"# Euler Problem\n",
"\n",
"2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.\n",
"\n",
"What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"My easiest guess is to multiply all prime numbers till the number."
]
},
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"source": [
"def get_primes_smaller(number):\n",
" primes = []\n",
" prospects = [n for n in range(2, number)]\n",
" while prospects:\n",
" p = prospects[0]\n",
" prospects = [x for x in prospects if x % p != 0]\n",
" primes.append(p)\n",
" return primes"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"210\n"
]
}
],
"source": [
"from operator import mul\n",
"from functools import reduce\n",
"\n",
"def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n",
" ps= get_primes_smaller(n + 1)\n",
" return reduce(mul, ps, 1)\n",
"\n",
"print(get_number_which_is_divisible_by_all_numbers_from_one_to(10))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"That obviously didn't work. The reason is that the same prime can occur multiple times in the factorization of a divisor. For example $2^{3} = 8$. We can always brute force of course. We do a smart brute force and only check multiples from the product of primes because this factor must be part of the solution."
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"232792560\n"
]
}
],
"source": [
"def is_divisible_by_numbers_smaller_or_equal(number, maximum_number):\n",
" for n in range(2, maximum_number + 1):\n",
" if number % n != 0:\n",
" return False\n",
" return True\n",
"\n",
"def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n",
" ps = get_primes_smaller(n + 1)\n",
" factor = reduce(mul, ps, 1)\n",
" multiples_of_factor = factor\n",
" while True:\n",
" if is_divisible_by_numbers_smaller_or_equal(multiples_of_factor, n):\n",
" return multiples_of_factor\n",
" multiples_of_factor += factor\n",
"\n",
"assert(get_number_which_is_divisible_by_all_numbers_from_one_to(10) == 2520)\n",
"print(get_number_which_is_divisible_by_all_numbers_from_one_to(20))"
]
}
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},
"tags": [
"reduce",
"divisible"
]
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