Sum multiples of 3 and 5: Difference between revisions
m
→Efficient solution
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# 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
Unfortunately, RPL can only handle double precision numbers, which prevents from precisely calculating the sum for n > 1E+15
≪ → n
≪ 0 1 3 FOR j
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