def main( prime_cap ):
#tries to find lowests sum set of 5 primes which all concatenate pair-wise to form primes
#only searches for primes below prime_cap
import import_primes
primes = import_primes.main()
#pulls all primes below one million
if prime_cap > ( 10 ** 6 ):
print('error, prime_cap > 10**6')
i = 0
while primes[ i ] < prime_cap and i < ( len( primes ) - 1 ):
i += 1
primes = primes[ 0 : i ]
#removes all primes above prime_cap from list of primes
good_sets = []
#will store all sets of primes found
for a in primes:
for b in primes:
if b > a:
if check_concat( b , a ):
for c in primes:
if c > b:
if check_concat( c , a ) and check_concat( c , b ):
for d in primes:
if d > c:
if check_concat( d , a ) and check_concat( d , b ) and check_concat( d , c ):
for e in primes:
if e > d:
if check_concat( e , a ) and check_concat( e , b ) and check_concat( e , c ) and check_concat( e , d ):
good_sets.append( [ a , b , c , d , e ] )
#TEST
print('good_sets=',good_sets)
#TEST
min_sum = sum( good_sets[ 0 ] )
stored_set = good_sets[ 0 ]
for good_set in good_sets:
if sum( good_set ) < min_sum:
min_sum = sum( good_set )
stored_set = good_set
return ( stored_set , min_sum )
def factor_below(x):
#finds prime factors of all numbers from 1 to x
primes = import_primes.main()
factor_list = [0, 1]
#zeroth and first entries are placeholders, after that i-th entry will hold factors of i
for i in range(2, x + 1):
#factors i and adds a set (unqiue entries) of its factors to the factor_list
factor_list.append(set(factor_given_primes.main(i, primes)))
#TEST
print(i)
#TEST
return factor_list
def main(x):
#finds number less than or equal to x (in problem x = one million) that contains the most prime factors
#this number solves the desired problem -- see wikipedia page
primes = import_primes.main()
#pulls list of primes less than one million
i = 0
num = 2
while num <= x:
#multiples number by primes until it is greater than x
i += 1
num = num * primes[i]
#multiply number by next prime
num = num // primes[i]
#divides out last prime number that was multiplied, to return number to below x
return num
#Prime summations #It is possible to write ten as the sum of primes in exactly five different ways: #7 + 3 #5 + 5 #5 + 3 + 2 #3 + 3 + 2 + 2 #2 + 2 + 2 + 2 + 2 #What is the first value which can be written as the sum of primes in over five thousand different ways? import import_primes primes_list = import_primes.main() #fetches list of all primes below one million import copy #for use later #COPIED BELOW FROM partition_function.py #WITH MINOR CHANGES ADDED #see: https://en.wikipedia.org/wiki/Partition_(number_theory) #NOTE: ASSUMES INPUT OF 7 OR GREATER import list_prod_sum #list_prod_sum.main(a,b) takes two lists a multiplies them pairwise. returns the sum. def rem(x, a, b):
#Prime square remainders
#Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)^n + (pn+1)^n is divided by pn^2.
#For example, when n = 3, p3 = 5, and 4^3 + 6^3 = 280 ≡ 5 mod 25.
#The least value of n for which the remainder first exceeds 10^9 is 7037.
#Find the least value of n for which the remainder first exceeds 10^10.
import import_primes
primes = import_primes.main()
#gets list of all primes below 1 million
def main(x):
#finds least value for n where remainder exceeds x (in problem as defined, x = 10^10)
i = 1
#prime index
p = 2
#first prime
rem = 1
#first remainder
while rem < x:
i += 1
p = primes[(i - 1)]
#pulls prime from list
rem = (((p - 1)**i) + ((p + 1)**i)) % (p**2)
