Problem 1

# Problem_1.py

"""

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

1000一下3的倍数以及5的倍数之和

"""

three = [i for i in range(0, 1000, 3)]

five = [i for i in range(0, 1000, 5)]

overlap = [i for i in three if i in five]

all = sum(three) + sum(five)

all -= sum(overlap)

print(all)

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