Resilience
Problem 243
A positive fraction whose numerator is less than its denominator is called a
proper fraction.
For any denominator, d, there will be d−1 proper fractions; for example, with
d = 12:
1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 , 8/12 , 9/12 , 10/12 , 11/12 .
We shall call a fraction that cannot be cancelled down a resilient fraction.
Furthermore we shall define the resilience of a denominator, R(d), to be the
ratio of its proper fractions that are resilient; for example, R(12) = 4/11 .
In fact, d = 12 is the smallest denominator having a resilience R(d) < 4/10 .
Find the smallest denominator d, having a resilience R(d) < 15499/94744 .
"""
from pe069 import prime_sieve
__date__ = '14-4-19'
__author__ = 'SUN'
if __name__ == '__main__':
prime = prime_sieve(30)
n = 1
d = 1
for i in range(len(prime)):
n *= prime[i] - 1
d *= prime[i]
for j in range(2, prime[i]):
if n * j / (d * j - 1) < 15499 / 94744:
print(d * j)
exit(0)
"""
Prime square remainders
Problem 123
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 time
from pe069 import prime_sieve
__date__ = '14-4-24'
__author__ = 'SUN'
if __name__ == '__main__':
start = time.clock()
prime = prime_sieve(300000)
for i in range(0, len(prime), 2):
if 2 * (i + 1) * prime[i] % prime[i] ** 2 > 10000000000:
print(i + 1)
break
print('Runtime is', time.clock() - start)
"""
Prime square remainders
Problem 123
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 time
from pe069 import prime_sieve
__date__ = '14-4-24'
__author__ = 'SUN'
if __name__ == '__main__':
start = time.clock()
prime = prime_sieve(300000)
for i in range(0, len(prime), 2):
if 2 * (i + 1) * prime[i] % prime[i]**2 > 10000000000:
print(i + 1)
break
print('Runtime is', time.clock() - start)
Resilience Problem 243 A positive fraction whose numerator is less than its denominator is called a proper fraction. For any denominator, d, there will be d−1 proper fractions; for example, with d = 12: 1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 , 8/12 , 9/12 , 10/12 , 11/12 . We shall call a fraction that cannot be cancelled down a resilient fraction. Furthermore we shall define the resilience of a denominator, R(d), to be the ratio of its proper fractions that are resilient; for example, R(12) = 4/11 . In fact, d = 12 is the smallest denominator having a resilience R(d) < 4/10 . Find the smallest denominator d, having a resilience R(d) < 15499/94744 . """ from pe069 import prime_sieve __date__ = '14-4-19' __author__ = 'SUN' if __name__ == '__main__': prime = prime_sieve(30) n = 1 d = 1 for i in range(len(prime)): n *= prime[i] - 1 d *= prime[i] for j in range(2, prime[i]): if n * j / (d * j - 1) < 15499 / 94744: print(d * j) exit(0)
'''
Summation of primes
Problem 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
'''
import time
from pe069 import prime_sieve
__author__ = 'SUN'
if __name__ == '__main__':
start = time.clock()
print(sum(prime_sieve(2000000)))
print('Runtime is', time.clock() - start)
phi = phi // i * (i - 1)
if n > 1:
phi *= n - 1
return phi
def totient_array(n):
array = [x for x in range(n + 1)]
for i in range(2, n + 1):
if array[i] == i:
for j in range(i, n + 1, i):
array[j] = array[j] // i * (i - 1)
return array
if __name__ == '__main__':
start = time.clock()
N = 10000000
prime = prime_sieve(4000)
answer = 0
ratio = 3
for i in range(len(prime)):
for j in range(i, len(prime)):
phi = prime[i] * prime[j]
if phi > N:
break
if sorted(str((prime[i] - 1) * (prime[j] - 1))) == sorted(str(phi)) \
and ratio > phi / ((prime[i] - 1) * (prime[j] - 1)):
answer = phi
ratio = phi / ((prime[i] - 1) * (prime[j] - 1))
print(answer)
print('Runtime is ', time.clock() - start)
'''
10001st prime
Problem 7
By listing the first six prime numbers: 2, 3,
5, 7, 11, and 13, we can see that the 6th prime
is 13. What is the 10 001st prime number?
'''
from pe069 import prime_sieve
__author__ = 'SUN'
if __name__ == '__main__':
print(prime_sieve(150000)[10000])
if n > 1:
phi *= n - 1
return phi
def totient_array(n):
array = [x for x in range(n + 1)]
for i in range(2, n + 1):
if array[i] == i:
for j in range(i, n + 1, i):
array[j] = array[j] // i * (i - 1)
return array
if __name__ == '__main__':
start = time.clock()
N = 10000000
prime = prime_sieve(4000)
answer = 0
ratio = 3
for i in range(len(prime)):
for j in range(i, len(prime)):
phi = prime[i] * prime[j]
if phi > N:
break
if sorted(str((prime[i] - 1) * (prime[j] - 1))) == sorted(str(phi)) \
and ratio > phi / ((prime[i] - 1) * (prime[j] - 1)):
answer = phi
ratio = phi / ((prime[i] - 1) * (prime[j] - 1))
print(answer)
print('Runtime is ', time.clock() - start)
