Project Euler Problem 5is one of those problems that seem hard, but turn out to be trivial after you think about them. In this case it is very important to use the information given to you in the problem. This means using the fact that 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. We can therefore limit our search to multiples of 2520.

**1. Create a divisors array**

First we will create a divisors array. We only need to divide by the numbers 11 – 20.

divisors = numpy.arange(11, 21) |

**2. Check for 0 remainder**

Second make sure that all the divisions of a number by the divisors produce a 0 remainder. We can check for that with the NumPy all function.

numpy.all((i % divisors) == 0) |

**3. Test the solution**

Eventually we will get an answer. We need to check whether this is the correct answer. Use the assert_equal function from the numpy.testing module to confirm the result.

numpy.testing.assert_equal(numpy.zeros(19), i % numpy.arange(2, 21)) |

Below is the complete code solution.

import numpy #2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. #What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? # 1. Create a divisors array divisors = numpy.arange(11, 21) for i in xrange(2520, 10 ** 9, 2520): # 2. Check for 0 remainder if numpy.all((i % divisors) == 0): print i # 3. Test the solution numpy.testing.assert_equal(numpy.zeros(19), i % numpy.arange(2, 21)) break |

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