Take the number 192 and multiply it by each of 1, 2, and 3:
192 × 1 = 192
192 × 2 = 384
192 × 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
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pans = []
for p in range(1,10**4):
digits = set([str(i) for i in range(1,10)])
n = 1
while len(digits) > 0:
pn = str(p*n)
d = set(pn)
if not(d.issubset(digits) and len(d) == len(pn)):break
n += 1
digits = digits.difference(d)
if len(digits) == 0:
pans.append(int(''.join([str(p*i) for i in range(1,n)])))
print "is pan:", p, "n:", n-1, pans[-1]
print "Max:", max(pans)
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