def is_special_number(numbers):
i = 1
for d in divisibleBy:
if permutations.combine_numbers(numbers[i : (i + 3)]) % d != 0:
return False
i += 1
return True
import primes
import permutations
primesToCheck = primes.create_primes(sqrt(987654321))
primeFound = False
def is_prime(number):
limit = sqrt(number)
for p in primesToCheck:
if p > limit:
return True
elif number % p == 0:
return False
return True
for x in xrange(10, 1, -1):
allPermutations = permutations.permutations(range(1, x))
for perm in sorted([permutations.combine_numbers(y) for y in allPermutations], reverse=True):
if is_prime(perm):
print str(perm)
primeFound = True
break
if primeFound:
break
192 x 1 = 192
192 x 2 = 384
192 x 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576.
We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645,
which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number
that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
"""
import permutations
allPermutations = set([permutations.combine_numbers(x) for x in permutations.permutations(range(1, 10))])
highestNumber = 0
for n in xrange(1, 10000): # the was found experimentally
pandigitalMultiply = 0
for i in xrange(1, 10):
pandigitalPart = n * i
pandigitalMultiply *= (10 ** len(str(pandigitalPart)))
pandigitalMultiply += pandigitalPart
if len(str(pandigitalMultiply)) >= 9:
break
if pandigitalMultiply in allPermutations:
if pandigitalMultiply > highestNumber:
- d4 d5 d6 = 635 is divisible by 5
- d5 d6 d7 = 357 is divisible by 7
- d6 d7 d8 = 572 is divisible by 11
- d7 d8 d9 = 728 is divisible by 13
- d8 d9 d10 = 289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
"""
import permutations
allPermutations = permutations.permutations(range(0, 10))
divisibleBy = [2, 3, 5, 7, 11, 13, 17]
sumOfPermutations = 0
def is_special_number(numbers):
i = 1
for d in divisibleBy:
if permutations.combine_numbers(numbers[i : (i + 3)]) % d != 0:
return False
i += 1
return True
for perm in allPermutations:
if is_special_number(perm):
sumOfPermutations += permutations.combine_numbers(perm)
print str(sumOfPermutations)
The number, 197, is called a circular prime because all rotations of the digits:
197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
"""
from collections import deque
import primes
import permutations
allPrimes = primes.create_primes(1000*1000)
countOfCircularPrimes = 0
for prime in allPrimes:
isCircularPrime = True
digits = deque(permutations.split_into_digits(prime))
for i in xrange(0, len(digits)):
digits.rotate(1)
newPrime = permutations.combine_numbers(digits)
if newPrime not in allPrimes:
isCircularPrime = False
break
if isCircularPrime:
countOfCircularPrimes += 1
print str(countOfCircularPrimes)
The product 7254 is unusual, as the identity, 39 x 186 = 7254,
containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
"""
import permutations
foundProducts = {}
foundProductsSum = 0
for perm in permutations.permutations(range(1, 10)):
product = permutations.combine_numbers(perm[-4:]) # last four digits
for x in xrange(1, 4):
factor1 = perm[:x] # first x digit
factor2 = perm[x:-4] # starting from x'th digit + 1 (0-based index) all except the last four
factor1 = permutations.combine_numbers(factor1)
factor2 = permutations.combine_numbers(factor2)
if factor1 * factor2 == product:
if product not in foundProducts:
foundProductsSum += product
foundProducts[product] = 1
#print str(factor1) + " x " + str(factor2) + " = " + str(product)
print str(foundProductsSum)
