def f(n, m, p):
# assert p in primes
return pow_mod(n, m % (p-1), p)
#!/usr/bin/python # -*- coding: utf-8 -*- #The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 26972593−1; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2p−1, have been found which contain more digits. #However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433×27830457+1. #Find the last ten digits of this prime number. #Answer: #8739992577 from time import time; t=time() from mathplus import pow_mod M = 10**10 print((28433*pow_mod(2, 7830457, M)) % M + 1)#, time()-t
def report(p):
print(p)#, time()-t)
import sys
sys.exit()
for p in range(S, T, 2):
if p % 5 == 0: continue
phip = phi(p)
m = p
if p % 3 == 0:
phip *= 9
m *= 9
if phip <= M: continue
factors = factorization(phip)
while phip > M:
for f, c in factors:
phif = phip//f
if pow_mod(10, phif, m) == 1:
phip = phif
if c > 1:
factors = [(k, v) if k != f else (k, v-1)
for k, v in factors]
else:
factors = [(k, v) for k, v in factors if k != f]
break
else:
report(p)
print('Failed!', time()-t)
#!/usr/bin/python
# -*- coding: utf-8 -*-
#The series, 11 + 22 + 33 + ... + 1010 = 10405071317.
#Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.
#Answer:
#9110846700
from mathplus import pow_mod
M = 1000
L = 10
p = 10**L
print(sum(pow_mod(i, i, p) for i in range(1, M)) % p)
#Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).
#Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).
#Answer:
#453647705
from time import time
t = time()
from mathplus import get_phis_by_sieve, pow_mod
M = 100000
phis = get_phis_by_sieve(M)
s = 3
for p, phip in enumerate(phis):
if phip != p - 1 or p == 1: continue
k = 1
while phip % 2 == 0:
phip //= 2
k *= 2
while phip % 5 == 0:
phip //= 5
k *= 5
if pow_mod(10, k, p) != 1:
s += p
print(s) #, time()-t)
def f(a, b, p):
if b == 1 or p == 2: return a % p
x = f(a, b-1, phi(p))
return pow_mod(a, x, p)
#Let us consider repunits of the form R(10n).
#Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).
#Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).
#Answer:
#453647705
from time import time; t=time()
from mathplus import get_phis_by_sieve, pow_mod
M = 100000
phis = get_phis_by_sieve(M)
s = 3
for p, phip in enumerate(phis):
if phip != p - 1 or p == 1: continue
k = 1
while phip % 2 == 0:
phip //= 2
k *= 2
while phip % 5 == 0:
phip //= 5
k *= 5
if pow_mod(10, k, p) != 1:
s += p
print(s)#, time()-t)
def f(a, b, p):
if b == 1 or p == 2: return a % p
x = f(a, b - 1, phi(p))
return pow_mod(a, x, p)
#You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
#However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
#Find the sum of the first twenty-five composite values of n for which
#GCD(n, 10) = 1 and n − 1 is divisible by A(n).
#Answer:
#149253
from time import time; t=time()
from mathplus import pow_mod, get_primes_by_sieve
M = 10**6
C = 25
L = []
primes, sieves = get_primes_by_sieve(M)
n, c = 7, C
while True:
n += 2
if n % 3 == 0 or n % 5 == 0 or sieves[n]: continue
if pow_mod(10, n-1, n) == 1:
c -= 1
L.append(n)
if c == 0: break
print(sum(L))#, time()-t
def f(n, m, p):
# assert p in primes
return pow_mod(n, m % (p - 1), p)
#!/usr/bin/python # -*- coding: utf-8 -*- # The series, 11 + 22 + 33 + ... + 1010 = 10405071317. # Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000. # Answer: # 9110846700 from mathplus import pow_mod M = 1000 L = 10 p = 10 ** L print(sum(pow_mod(i, i, p) for i in range(1, M)) % p)
#!/usr/bin/python # -*- coding: utf-8 -*- #The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 26972593−1; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2p−1, have been found which contain more digits. #However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433×27830457+1. #Find the last ten digits of this prime number. #Answer: #8739992577 from time import time t = time() from mathplus import pow_mod M = 10**10 print((28433 * pow_mod(2, 7830457, M)) % M + 1) #, time()-t
#You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
#However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
#Find the sum of the first twenty-five composite values of n for which
#GCD(n, 10) = 1 and n − 1 is divisible by A(n).
#Answer:
#149253
from time import time
t = time()
from mathplus import pow_mod, get_primes_by_sieve
M = 10**6
C = 25
L = []
primes, sieves = get_primes_by_sieve(M)
n, c = 7, C
while True:
n += 2
if n % 3 == 0 or n % 5 == 0 or sieves[n]: continue
if pow_mod(10, n - 1, n) == 1:
c -= 1
L.append(n)
if c == 0: break
print(sum(L)) #, time()-t
