Sum multiples of 3 and 5: Difference between revisions
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===Efficient solution===
This is a straightforward way 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]
# 1+2+...k = k*(k+1)/2
≪ → n
≪ 0 1 3 FOR j
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