# Author: Deddryk
"""
Solution to problem 38
"""
from utils.misc import used_digits, num_digits
def is_pandigital(n):
return len(used_digits(n)) == num_digits(n)
concatenated_pandigitals = set()
for i in xrange(10, 10000):
multiplier = 1
concatenated_product = 0
while num_digits(concatenated_product) < 9:
next_part = i * multiplier
multiplier += 1
concatenated_product = concatenated_product * 10**num_digits(
next_part) + next_part
if not is_pandigital(concatenated_product): break
if 0 in used_digits(concatenated_product): break
else:
if num_digits(concatenated_product) == 9:
concatenated_pandigitals.add(concatenated_product)
print max(concatenated_pandigitals)
def is_pandigital(n):
return len(used_digits(n)) == num_digits(n)
def is_pandigital(n):
return len(used_digits(n)) == num_digits(n)
# Author: Deddryk
"""
Solution to problem 38
"""
from utils.misc import used_digits, num_digits
def is_pandigital(n):
return len(used_digits(n)) == num_digits(n)
concatenated_pandigitals = set()
for i in xrange(10, 10000):
multiplier = 1
concatenated_product = 0
while num_digits(concatenated_product) < 9:
next_part = i * multiplier
multiplier += 1
concatenated_product = concatenated_product * 10 ** num_digits(next_part) + next_part
if not is_pandigital(concatenated_product):
break
if 0 in used_digits(concatenated_product):
break
else:
if num_digits(concatenated_product) == 9:
concatenated_pandigitals.add(concatenated_product)
print max(concatenated_pandigitals)
Solution to problem 32.
This solution is ugly but it works quickly. Using sets I limit the search space
to only those numbers that do not have similar digits, and only 3 or 4 digits.
"""
from math import log10, floor
from utils.misc import used_digits
digits = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
m1 = set()
m2 = set()
products = set()
for i in digits:
for j in digits - used_digits(i):
for k in digits - used_digits(i) - used_digits(j):
m1.add(100*i + 10*j + k)
for l in digits - used_digits(i) - used_digits(j) - used_digits(k):
m2.add(1000*i + 100*j + 10*k + l)
for i in m1:
m3 = set()
for j in digits - used_digits(i):
for k in digits - used_digits(i) - used_digits(j):
m3.add(10 * j + k)
for j in m3:
if floor(log10(i*j)) == 3 and len(digits - used_digits(i) - used_digits(j) - used_digits(i*j)) == 0:
products.add(i*j)
for i in m2:
for j in digits - used_digits(i):
Solution to problem 32.
This solution is ugly but it works quickly. Using sets I limit the search space
to only those numbers that do not have similar digits, and only 3 or 4 digits.
"""
from math import log10, floor
from utils.misc import used_digits
digits = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
m1 = set()
m2 = set()
products = set()
for i in digits:
for j in digits - used_digits(i):
for k in digits - used_digits(i) - used_digits(j):
m1.add(100 * i + 10 * j + k)
for l in digits - used_digits(i) - used_digits(j) - used_digits(k):
m2.add(1000 * i + 100 * j + 10 * k + l)
for i in m1:
m3 = set()
for j in digits - used_digits(i):
for k in digits - used_digits(i) - used_digits(j):
m3.add(10 * j + k)
for j in m3:
if floor(log10(
i * j)) == 3 and len(digits - used_digits(i) - used_digits(j) -
used_digits(i * j)) == 0:
products.add(i * j)
"""
from utils.misc import used_digits, gcd
from utils.primes import factor
num = 1
den = 1
def check_eq(a, b, c, d, num, den):
if a % 10 == 0 and b % 10 == 0: return num, den
if a / gcd(a, b) == c / gcd(c, d) and b / gcd(a, b) == d / gcd(c, d):
num *= a
den *= b
return num, den
for i in xrange(10, 100):
for j in xrange(i + 1, 100):
if len(used_digits(i) & used_digits(j)) > 0:
if i / 10 == j / 10:
num, den = check_eq(i, j, i % 10, j % 10, num, den)
elif i / 10 == j % 10:
num, den = check_eq(i, j, i % 10, j / 10, num, den)
elif i % 10 == j / 10:
num, den = check_eq(i, j, i / 10, j % 10, num, den)
else:
num, den = check_eq(i, j, i / 10, j / 10, num, den)
print den / gcd(num, den)
