def compute():
valid_pandigitals = []
for x in range(1, 10):
concat = str(x) + str(x * 2) + str(x * 3) + str(x * 4) + str(x * 5)
if is_pandigital(concat):
valid_pandigitals.append(concat)
for x in range(10, 100):
concat = str(x) + str(x * 2) + str(x * 3) + str(x * 4)
if is_pandigital(concat):
valid_pandigitals.append(concat)
for x in range(100, 1000):
concat = str(x) + str(x * 2) + str(x * 3)
if is_pandigital(concat):
valid_pandigitals.append(concat)
for x in range(1000, 10000):
concat = str(x) + str(x * 2)
if is_pandigital(concat):
valid_pandigitals.append(concat)
return max(valid_pandigitals)
def first_digits():
n, m = 1, 1
poss_k1 = []
for k in range(3, 5 * 10**5):
l = int(str(n + m)[:20])
if is_pandigital(str(l)):
poss_k1.append(k)
n, m = m, l
return poss_k1
def last_digits():
n, m = 1, 1
poss_k2 = []
for k in range(3, 5 * 10**5):
l = int(str(n + m)[-10:])
if is_pandigital(str(l)):
poss_k2.append(k)
n, m = m, l
return poss_k2
def find_largest_concatenation_of_products():
lower_bound = 918273645
upper_bound = 987654321
for n in range(upper_bound, lower_bound, -1):
if not is_pandigital(n):
continue
if is_concatenation_of_products(n):
return n
def is_expression_pandigital(m1, m2, p):
'''
The digits 1 to 9 must be included in the expression m1 * m2 = p
exactly once each
'''
return is_pandigital("%s%s%s" % (m1, m2, p))