#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.
#Answer:
#18613426663617118
#https://github.com/skydark/project-euler/blob/master/134.py
from time import time; t=time()
from mathplus import log10, get_primes_by_sieve
M = 10**6+5
primes, _ = get_primes_by_sieve(M)
def f(n, m):
#assert n > m
if (n, m) == (1, 0): return 1, 0
q, r = n//m, n%m
a, b = f(m, r)
return b, a-b*q
s = 0
for i in range(2, len(primes)-1):
p, q = primes[i:i+2]
n = 10**(int(log10(p))+1)
a, b = f(q, n%q)
s += (b*(q-p) % q)*n+p
print(s)#, time()-t
def concat(i, j):
return 10**(int(log10(j))+1)*i+j
#Find ∑ S for every pair of consecutive primes with 5 ≤ p1 ≤ 1000000.
#Answer:
#18613426663617118
#https://github.com/skydark/project-euler/blob/master/134.py
from time import time
t = time()
from mathplus import log10, get_primes_by_sieve
M = 10**6 + 5
primes, _ = get_primes_by_sieve(M)
def f(n, m):
#assert n > m
if (n, m) == (1, 0): return 1, 0
q, r = n // m, n % m
a, b = f(m, r)
return b, a - b * q
s = 0
for i in range(2, len(primes) - 1):
p, q = primes[i:i + 2]
n = 10**(int(log10(p)) + 1)
a, b = f(q, n % q)
s += (b * (q - p) % q) * n + p
print(s) #, time()-t
#So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15.
#There are 1105 Hamming numbers not exceeding 108.
#We will call a positive number a generalised Hamming number of type n, if it has no prime factor larger than n.
#Hence the Hamming numbers are the generalised Hamming numbers of type 5.
#How many generalised Hamming numbers of type 100 are there which don't exceed 109?
#Answer:
#2944730
from time import time; t=time()
from mathplus import log10
M = 9
P = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
pl = [log10(p) for p in P]
def f(pos, limit):
if pos == 0: return int(limit/pl[0])+1
v = pl[pos]
pos -= 1
s = 0
while limit > 0:
s += f(pos, limit)
limit -= v
return s
print(f(len(pl)-1, M))#, time()-t
