import timing
import eulerutil
import itertools
import copy
'''
Circular primes
Problem 35
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?
'''
def rotations(n):
results = []
num = [str(i) for i in str(n)]
for i in range(len(num)):
num.insert(0,num.pop())
results.append(''.join(list(num)))
return results
max = 1000000
count = 0
for n in range(2,max):
if eulerutil.isprime(n):
if all([eulerutil.isprime(int(i)) for i in rotations(n)]):
count += 1
if n%10000==0: print(n)
print(count)
The incredible formula n² − 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, −79 and 1601, is −126479. Considering quadratics of the form: n² + an + b, where |a| < 1000 and |b| < 1000 where |n| is the modulus/absolute value of n e.g. |11| = 11 and |−4| = 4 Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0. ''' def f(n,a,b): return (n**2)+(a*n)+b max = 0 for a in range(-999,1000): for b in range(-999,1000): n = 0 while True: if eulerutil.isprime(f(n,a,b)): n+=1 else: break if n > max: max = n prod = a*b print(max,prod)
The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3. Find the sum of the only eleven primes that are both truncatable from left to right and right to left. NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes. ''' def truncate(n): num = str(n) result = [num] for i in range(1,len(num)): result.append(num[:i]) result.append(num[i:]) return [int(x) for x in result] count, i = 0, 11 truncatable_primes = [] while True: if count>=11: break if isprime(i): if all([isprime(x) for x in truncate(i)]): count += 1 truncatable_primes.append(i) i+=1 if i%100000==0: print(i,count) print(truncatable_primes) print(sum(truncatable_primes))
import timing from eulerutil import isprime from itertools import permutations ''' Pandigital prime Problem 41 We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime. What is the largest n-digit pandigital prime that exists? ''' digits = "123456789" p = [] for n in range(3,10): for i in [''.join(list(x)) for x in permutations(digits[:n],n)]: if isprime(int(i)): p+=[i] print(max(p))
