def p46():
#getting precalculated primes and squares:
u = Utils()
primes = u.sieve(6000)
squares = [i * i for i in range(1, 101)]
#start searching at 15:
o = 15
while o < 5800:
#is the test number prime?
if u.chop(o, primes) != -1:
#yeap, go to next one:
o += 2
continue
#nope, begin testing:
ind = 0
found = False
while o - primes[ind] > 0:
s = (o - primes[ind]) / 2
#found desired decomposition?
if u.chop(s, squares) != -1:
#yep, break out:
found = True
break
#nope, try next prime:
else:
ind += 1
#found possible candidate:
if not found:
return o
#keep searching:
o += 2
def p231():
u = Utils()
primes = u.sieve(20000000)
total = 0
for p in primes:
total += p * (f.v(20000000, p) - (f.v(5000000, p) + f.v(15000000, p)))
return total
def p35():
u = Utils()
sieve = u.sieve(10**6)
count = 0
for prime in sieve:
s = str(prime)
l = cyclic_shifts_of(s)
if all_in(l, sieve, u):
count += 1
return count
# factors. For example, 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2
# × 3.
#
# There are ten composites below thirty containing precisely
# two, not necessarily distinct, prime factors: 4, 6, 9, 10,
# 14, 15, 21, 22, 25, 26.
#
# How many composite integers, n < 10^8, have precisely two,
# not necessarily distinct, prime factors?
from euler.utils import Utils
from math import sqrt
from bisect import bisect
u = Utils()
#One of the accepted solutions from the forum (by
#logopetria, @Sat, 22 Mar 2008, 08:45):
primes = u.sieve(5 * (10**7))
def p187():
N = 10**8
total = 0
for x in range(bisect(primes, sqrt(N))):
p = primes[x]
total += bisect(primes, N / p) - x
return total
print(p187())
# Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that
# 1219 is the smallest number such that the last digits are formed by p1 whilst
# also being divisible by p2.
#
# In fact, with the exception of p1 = 3 and p2 = 5, for every pair of consecutive
# primes, p2 > p1, there exist values of n for which the last digits are formed by
# p1 and n is divisible by p2. Let S be the smallest of these values of n.
#
# Find S for every pair of consecutive primes with 5 <= p1 <= 1000000.
from euler.utils import Utils
u = Utils()
a = u.sieve(1100000)
def phi(n):
"""
It is assumed that n is a power of 10.
"""
return (2 * n) // 5
def f1(p1, p2):
m = len(str(p1))
r = 10 ** m
q = phi(r) - 1
s = pow(p2, q, r)
return ((((p1 * p2 * s) // r) % p2) * r) + p1
def p134():
l = [f1(a[i], a[i + 1]) for i in range(2, len(a) - 1) if a[i] < 10 ** 6]
return sum(l)
#coding: UTF-8
# A composite is a number containing at least two prime
# factors. For example, 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2
# × 3.
#
# There are ten composites below thirty containing precisely
# two, not necessarily distinct, prime factors: 4, 6, 9, 10,
# 14, 15, 21, 22, 25, 26.
#
# How many composite integers, n < 10^8, have precisely two,
# not necessarily distinct, prime factors?
from euler.utils import Utils
from math import sqrt
from bisect import bisect
u = Utils()
#One of the accepted solutions from the forum (by
#logopetria, @Sat, 22 Mar 2008, 08:45):
primes = u.sieve(5 * (10 ** 7))
def p187():
N=10**8
total=0
for x in range(bisect(primes, sqrt(N))):
p = primes[x]
total += bisect(primes, N/p) - x
return total
print(p187())
def test_sieve():
u = Utils()
primes = u.sieve(50)
assert(primes == [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47])
# is strongly advised that you solve this one first.
from euler.utils import Utils
u = Utils()
MAX = 200000
def get_exp(n, p):
i = 0
while n % p == 0:
i += 1
n = n / p
return i
prime_list = u.sieve(MAX)
# calculating the desired number using memoization, stopping
# as soon as we find it:
def p108():
i = 2
l = [0, 1]
while i < MAX:
for p in prime_list:
if i % p == 0:
val = l[i // p] + \
2 * l[i // (p ** get_exp(i, p))] - 1
if val >= 1000:
return i
else:
l.append(val)
# square.
#
# Given that the frog's starting position is random with the same probability for
# every square, and given that she listens to his first 15 croaks, what is the
# probability that she hears the sequence PPPPNNPPPNPPNPN?
#
# Give your answer as a fraction p/q in reduced form.
from fractions import Fraction
from euler.utils import Utils
u = Utils()
d = {}
primes = u.sieve(500)
for i in range(1, 501):
if i in primes:
d[("P", i)] = Fraction(2, 3)
d[("N", i)] = Fraction(1, 3)
else:
d[("N", i)] = Fraction(2, 3)
d[("P", i)] = Fraction(1, 3)
def p(s, i):
if (s, i) in d.keys():
return d[(s, i)]
else:
if i == 1:
# a perfect cube.
#
# For example, when p = 19, 8^3 + 8^2×19 = 12^3.
#
# What is perhaps most surprising is that for each prime with
# this property the value of n is unique, and there are only
# four such primes below one-hundred.
#
# How many primes below one million have this remarkable
# property?
from euler.utils import Utils
cube_list = [x**3 for x in range(578)]
u = Utils()
prime_list = u.sieve(10**6)
def p131():
"""
As taken from the forum:
Since x^3 = n^2(n + p), and p is a prime, it turns
out that n must be a cube, as well as n + p, i.e.,
we must have p = a^3 - b^3 for some a, b.
But, for a^3 - b^3 = (a - b)(a^2 + ab + b^2) to be
prime we must have a - b = 1, so p must be a
difference of consecutive cubes.
"""
total = 0
for i in range(len(cube_list) - 1):
#coding: UTF-8
# Let p_n be the nth prime: 2, 3, 5, 7, 11, ..., and let r
# be the remainder when (p_n − 1)^n + (p_n + 1)^n is divided
# by p_n^2.
#
# For example, when n = 3, p_3 = 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.
from euler.utils import Utils
u = Utils()
p = u.sieve(4 * (10 ** 5))
def p123():
i = 0
a = 0
while a < 10 ** 10 and i < len(p):
i += 1
a = (2 * p[i] * (i + 1)) % (p[i] ** 2)
# sum 2 because i is the index of the number just below
# 10 ** 10, and array indices start by 0:
return i + 2
print(p123())
# be the remainder when (p_n − 1)^n + (p_n + 1)^n is divided
# by p_n^2.
#
# For example, when n = 3, p_3 = 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.
from euler.utils import Utils
u = Utils()
p = u.sieve(4 * (10**5))
def p123():
i = 0
a = 0
while a < 10**10 and i < len(p):
i += 1
a = (2 * p[i] * (i + 1)) % (p[i]**2)
# sum 2 because i is the index of the number just below
# 10 ** 10, and array indices start by 0:
return i + 2
print(p123())
# square.
#
# Given that the frog's starting position is random with the same probability for
# every square, and given that she listens to his first 15 croaks, what is the
# probability that she hears the sequence PPPPNNPPPNPPNPN?
#
# Give your answer as a fraction p/q in reduced form.
from fractions import Fraction
from euler.utils import Utils
u = Utils()
d = {}
primes = u.sieve(500)
for i in range(1, 501):
if i in primes:
d[("P", i)] = Fraction(2, 3)
d[("N", i)] = Fraction(1, 3)
else:
d[("N", i)] = Fraction(2, 3)
d[("P", i)] = Fraction(1, 3)
def p(s, i):
if (s, i) in d.keys():
return d[(s, i)]
else:
if i == 1:
d[(s, 1)] = p(s[0], 1) * p(s[1:], 2)
# Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that
# 1219 is the smallest number such that the last digits are formed by p1 whilst
# also being divisible by p2.
#
# In fact, with the exception of p1 = 3 and p2 = 5, for every pair of consecutive
# primes, p2 > p1, there exist values of n for which the last digits are formed by
# p1 and n is divisible by p2. Let S be the smallest of these values of n.
#
# Find S for every pair of consecutive primes with 5 <= p1 <= 1000000.
from euler.utils import Utils
u = Utils()
a = u.sieve(1100000)
def phi(n):
"""
It is assumed that n is a power of 10.
"""
return (2 * n) // 5
def f1(p1, p2):
m = len(str(p1))
r = 10**m
q = phi(r) - 1
s = pow(p2, q, r)
return ((((p1 * p2 * s) // r) % p2) * r) + p1
