Here's my solution to Project Euler problem 6:

# where the numbers list ends n = 100 # handy function to square a number pow2 = lambda x: pow(x, 2) # first compute the sum of the squares sumsq = sum(map(pow2, range(1, n+1))) # then the square of the sum sqsum = pow2(sum(range(1, n+1))) print sqsum - sumsq

## 4 comments:

Why pow(x, 2) instead of x ** 2 or x * x?

@Marius: no real reason, I wanted to use map() :)

2 * sum([i*j for i in range(1, 101) for j in range(i+1, 101)]

untested, however the idea is that the difference contains each product of two integers exactly twice.

the benefit of this method is that intermediate results are not greater that the final result (hence there is no risk of overflow)

Even better, using some classic math results:

1+2...+n = n * (n+1)/2

1²+2²+...+n² = n*(n+1)*(2n+1)/6

do the difference and you'ĺl find:

n(n+1)(n-1)(3n+2)/12

@mpomme: that's what you get for working in the industry for such a long time: you forget the basic math - the series expansion solution is much better than doing all the calculation & summing

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