EularProject 38：数字组合1-9分解为数字与1到n的乘积的和

Pandigital multiples

Problem 38

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?

Answer:

932718654

Completed on Thu, 30 Jul 2015, 04:33

__author__ = ‘zhengyi‘

def IsPass(x):
    Nstr=str(x)
    k=1
    while len(Nstr)<9:
        k+=1
        Nstr+=str(k*x)
    if len(Nstr)>9:
        return None
    if len(set(Nstr))<9:
        return None
    if ‘0‘ in Nstr:
        return None
    else:
        return Nstr

result=‘‘
for i in range(10,pow(10,4)):
    temp=IsPass(i)
    if temp:
        result=temp

print(result)

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