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Line 3,976: ≫ ===Efficient solution=== This is a ~~straightforward~~fast ~~way~~approach to calculate the sum for higher values of n, taking into account that: # the sum of multiples of 3 and 5 is the sum of multiples of 3 plus the sum of multiples of 5, minus the sum of multiples of 15 to remove double counting # the sum of multiples of m being < n is equal to m*(1+2+ ... k) with k =[n/m]

Sum multiples of 3 and 5: Difference between revisions

Sum multiples of 3 and 5 (view source)