def gen(l, m, i):
others = list(range(10))
others.remove(i)
past = set()
ll = [True]*m+[False]*(l-m)
for u in permutations(ll):
if u in past: continue
past.add(u)
if u[-1] and i % 2 == 0: continue
for v in product(*([others]*(l-m))):
if not u[-1] and v[-1] % 2 == 0: continue
n, k = 0, 0
for j in u:
n *= 10
if j:
n += i
else:
n += v[k]
k += 1
yield n
def get_ret3(abc):
ss = set()
for i in permutations(abc):
a, b, c = i[0], i[1], i[2]
for r in get_ret2(a, b):
ss.add(tuple(sorted((r, c))))
return set(r for ab in ss for r in get_ret2(*ab))
def test_xy(x, y):
global max_value
for f in permutations(range(10), len_active_k):
for j in range(len_active_k):
full_f[active_k[j]] = f[j]
if test(x) > 0 and test(y) > 0:
max_value = max(max_value, test(x), test(y))
def test_xy(x, y):
global max_value
for f in permutations(range(10), len_active_k):
for j in range(len_active_k):
full_f[active_k[j]] = f[j]
if test(x) > 0 and test(y) > 0:
max_value = max(max_value, test(x), test(y))
def gen(l, m, i):
others = list(range(10))
others.remove(i)
past = set()
ll = [True] * m + [False] * (l - m)
for u in permutations(ll):
if u in past: continue
past.add(u)
if u[-1] and i % 2 == 0: continue
for v in product(*([others] * (l - m))):
if not u[-1] and v[-1] % 2 == 0: continue
n, k = 0, 0
for j in u:
n *= 10
if j:
n += i
else:
n += v[k]
k += 1
yield n
#What 12-digit number do you form by concatenating the three terms in this sequence?
#Answer:
#296962999629
from time import time
t = time()
from mathplus import sieve, permutations
ret = {}
primes = sieve(10000)
p = [i for i, f in enumerate(primes) if f and i >= 1000]
for n in p:
c = []
for i in permutations(
(n // 1000, (n % 1000) // 100, (n % 100) // 10, n % 10)):
if 0 in i: break
m = i[0] * 1000 + i[1] * 100 + i[2] * 10 + i[3]
if primes[m] and m not in c: c.append(m)
if len(c) >= 3:
c = sorted(c)
if c[0] in ret: continue
for i in range(1, len(c) - 1):
for j in range(i):
if c[i] * 2 - c[j] in c:
ret[c[0]] = (c[j], c[i], 2 * c[i] - c[j])
assert len(ret) == 2
del ret[1487]
print(['%s%s%s' % (v[0], v[1], v[2]) for k, v in ret.items()][0]) #, time()-t
#!/usr/bin/python
# -*- coding: utf-8 -*-
#We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
#What is the largest n-digit pandigital prime that exists?
#Answer:
#7652413
from time import time
t = time()
from mathplus import first_factor, permutations, reduce
p = 0
for ps in permutations(range(7, 0, -1)):
if ps[0] == 5: continue
p = reduce(lambda x, y: 10 * x + y, ps)
if first_factor(p) == p:
break
print(p) #, time()-t
#Answer:
#16695334890
from time import time
t = time()
from mathplus import permutations, reduce
def three(d, i):
return d[i] * 100 + d[i + 1] * 10 + d[i + 2]
ret = []
for d in permutations(range(10)):
if d[5] != 0 and d[5] != 5: continue
if d[3] % 2 != 0: continue
#if three(d, 1) % 2 != 0: continue
if (d[2] + d[3] + d[4]) % 3 != 0: continue
#if three(d, 2) % 3 != 0: continue
#if three(d, 3) % 5 != 0: continue
if (d[4] * 2 + d[5] * 3 + d[6]) % 7 != 0: continue
#if three(d, 4) % 7 != 0: continue
if (d[5] - d[6] + d[7]) % 11 != 0: continue
#if three(d, 5) % 11 != 0: continue
if (d[6] * 9 - d[7] * 3 + d[8]) % 13 != 0: continue
#if three(d, 6) % 13 != 0: continue
if (-2 * d[7] - 7 * d[8] + d[9]) % 17 != 0: continue
#if three(d, 7) % 17 != 0: continue
ret.append(reduce(lambda x, y: 10 * x + y, d))
if a != 0: r.append(b/a)
if b != 0: r.append(a/b)
return r
@memorize
def get_ret3(abc):
ss = set()
for i in permutations(abc):
a, b, c = i[0], i[1], i[2]
for r in get_ret2(a, b):
ss.add(tuple(sorted((r, c))))
return set(r for ab in ss for r in get_ret2(*ab))
for k in combinations(range(1, 10), 4):
ss = set()
for i in permutations(k):
a, b, c, d = i[0], i[1], i[2], i[3]
for r in get_ret2(a, b):
ss.add(tuple(sorted((r, c, d))))
ret = set(r for abc in ss for r in get_ret3(abc))
i = 1
while True:
if i not in ret: break
i += 1
i -= 1
if i > s:
s = i
abcd = sorted(k)
print(''.join(str(i) for i in abcd))#, s, time()-t)
def direct_no_thinking(n, size):
return list(permutations(range(size)))[n]
#!/usr/bin/python
# -*- coding: utf-8 -*-
#We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
#What is the largest n-digit pandigital prime that exists?
#Answer:
#7652413
from time import time; t = time()
from mathplus import first_factor, permutations, reduce
p = 0
for ps in permutations(range(7, 0, -1)):
if ps[0] == 5: continue
p = reduce(lambda x, y: 10*x+y, ps)
if first_factor(p) == p:
break
print(p)#, time()-t
#* d7d8d9=728 is divisible by 13
#* d8d9d10=289 is divisible by 17
#Find the sum of all 0 to 9 pandigital numbers with this property.
#Answer:
#16695334890
from time import time; t=time()
from mathplus import permutations, reduce
def three(d, i):
return d[i]*100+d[i+1]*10+d[i+2]
ret = []
for d in permutations(range(10)):
if d[5] != 0 and d[5] != 5: continue
if d[3] % 2 != 0: continue
#if three(d, 1) % 2 != 0: continue
if (d[2]+d[3]+d[4]) % 3 != 0: continue
#if three(d, 2) % 3 != 0: continue
#if three(d, 3) % 5 != 0: continue
if (d[4]*2+d[5]*3+d[6]) % 7 != 0: continue
#if three(d, 4) % 7 != 0: continue
if (d[5]-d[6]+d[7]) % 11 != 0: continue
#if three(d, 5) % 11 != 0: continue
if (d[6]*9-d[7]*3+d[8]) % 13 != 0: continue
#if three(d, 6) % 13 != 0: continue
if (-2*d[7]-7*d[8]+d[9]) % 17 != 0: continue
#if three(d, 7) % 17 != 0: continue
ret.append(reduce(lambda x, y:10*x+y, d))
primes, sieves = get_primes_by_sieve(M)
def is_prime_plus(n):
if n % 2 == 0: return n == 2
if n % 3 == 0: return n == 3
if n < M: return sieves[n]
isqrtn = isqrt(n)
for p in primes:
if p > isqrtn: return True
if n % p == 0: return False
ret = set()
def tryn(n, l, start_i, tr, maxn):
for i in range(start_i, 9):
n = n*10+l[i]
if n < maxn: continue
if is_prime_plus(n):
tr.append(n)
tryn(0, l, i+1, tr, n)
tr.pop()
if start_i == 9:
ret.add(tuple(sorted(tr)))
for l in permutations(range(1, 10)):
if l[-1] % 2 == 0 or l[-1] == 5: continue
tr = []
tryn(0, l, 0, tr, 0)
print(len(ret))#, time()-t
#We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
#The product 7254 is unusual, as the identity, 39 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
#Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
#HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
#Answer:
#45228
from time import time; t=time()
from mathplus import permutations, reduce
def to_int(l):
return reduce(lambda x, y: x*10+y, l)
ret = set()
for i in permutations(range(1, 10), 9):
if i[0] >= 8: break
if i[0]*i[2] < 10 and i[1]*i[4] % 10 == i[8]:
r = to_int(i[5:])
if to_int(i[:2]) * to_int(i[2:5]) == r:
ret.add(r)
if i[0]*i[1] < 10 and i[0]*i[4] % 10 == i[8]:
r = to_int(i[5:])
if to_int(i[:1]) * to_int(i[1:5]) == r:
ret.add(r)
print(sum(ret))#, time()-t
def direct_no_thinking(n, size):
return list(permutations(range(size)))[n]
#There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
#What 12-digit number do you form by concatenating the three terms in this sequence?
#Answer:
#296962999629
from time import time; t=time()
from mathplus import sieve, permutations
ret = {}
primes = sieve(10000)
p = [i for i, f in enumerate(primes) if f and i >= 1000]
for n in p:
c = []
for i in permutations((n//1000, (n % 1000)//100, (n % 100)//10, n%10)):
if 0 in i: break
m = i[0]*1000+i[1]*100+i[2]*10+i[3]
if primes[m] and m not in c: c.append(m)
if len(c) >= 3:
c = sorted(c)
if c[0] in ret: continue
for i in range(1, len(c)-1):
for j in range(i):
if c[i]*2-c[j] in c:
ret[c[0]] = (c[j], c[i], 2*c[i]-c[j])
assert len(ret) == 2
del ret[1487]
print(['%s%s%s' %(v[0],v[1],v[2]) for k, v in ret.items()][0])#, time()-t
