def main(verbose=False):
for n in range(10):
if not check_formula(n):
raise Exception("Proposed formula for M(2**k + 1) incorrect.")
modulus = 7 ** 9
p_2 = 4 * modular_exponentiate(2, 10, 18, modulus) - 2
p_3 = 3 * modular_exponentiate(3, 10, 18, modulus) - 1
p_4 = 4 * modular_exponentiate(4, 10, 18, modulus) - 1
return (p_4 * inverse_mod_n(3, modulus) -
p_3 * inverse_mod_n(2, modulus) + p_2) % (modulus)
def main(verbose=False):
PRIMES = sieve(10 ** 7)
running_sum = 0
for prime in PRIMES:
if prime not in [2, 5]:
running_sum += inverse_mod_n(10, prime)
return running_sum
def main(verbose=False):
PRIMES = sieve(10 ** 6 + 3) # 10**6 + 3 is the final value of p_2
running_sum = 0
for index in range(2, len(PRIMES) - 1):
p_1 = PRIMES[index]
p_2 = PRIMES[index + 1]
ten_inverse = inverse_mod_n(10, p_2)
digits = len(str(p_1))
k = (ten_inverse ** digits) * (p_2 - p_1) % p_2
running_sum += int('%s%s' % (k, p_1))
return running_sum
def main(verbose=False):
p_3, c, P_2 = sequence('UDDDUdddDDUDDddDdDddDDUDDdUUDd', 0, 0, 1)
# Here a_1 = s**(-1)(y) = ((3**(p_3))*y + c)/P_2
# Since we need a_1 > 10**15, (3**(p_3))*y > (10**15)*P_2 - c
min_y = ((10 ** 15) * P_2 - c) / (3 ** p_3)
# We also need (3**(p_3))*y == -c mod P_2
y_residue = inverse_mod_n(3 ** (p_3), P_2) * (-c) % P_2
# integer division intended to find nearest multiple of P_2
y = P_2 * (min_y / P_2) + y_residue
if y < min_y:
y += P_2
return ((3 ** p_3) * y + c) / P_2
def last5(n):
if n < 8:
# The remaining part is always divisible by
# 2**5 for n > 7, these are special cases
solutions = {0: 1,
1: 1,
2: 2,
3: 6,
4: 24,
5: 12,
6: 72,
7: 504}
return solutions[n]
if n <= 5 ** 5:
residues = {}
for i in range(1, n + 1):
to_add = robust_divide(i, 5)
# if to_add is not in residues, sets to 1
# (default 0 returned by get)
residues[to_add] = residues.get(to_add, 0) + 1
else:
residues = {}
for residue in range(1, 5 ** 5):
if residue % 5 != 0:
residues[residue] = (n - residue) / (5 ** 5) + 1
max_power = int(floor(log(n) / log(5)))
for power in range(1, max_power + 1):
biggest_quotient = n / (5 ** power)
for residue in range(1, 5 ** 5):
if residue % 5 != 0:
residues[residue] += ((biggest_quotient - residue) /
(5 ** 5) + 1)
product = 1
for residue, power in residues.items():
power_apply = power % (4 * (5 ** 4)) # PHI(5**5)
product = (product * (residue ** power_apply)) % (5 ** 5)
fives = num_factors_fact(n, 5) % (4 * (5 ** 4)) # PHI(5**5)
inverse = inverse_mod_n(2, 5 ** 5)
product = (product * (inverse ** fives)) % (5 ** 5)
return (product * unit_a_zero_b(5 ** 5, 2 ** 5)) % 10 ** 5