Sum multiples of 3 and 5: Difference between revisions
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# the sum of multiples of m being < n is equal to m*(1+2+ ... k) with k =[n/m] |
# the sum of multiples of m being < n is equal to m*(1+2+ ... k) with k =[n/m] |
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# 1+2+...k = k*(k+1)/2 |
# 1+2+...k = k*(k+1)/2 |
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Unfortunately, RPL can only handle double precision numbers, which prevents from precisely calculating the sum for n > 1E+15 |
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≪ → n |
≪ → n |
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≪ 0 1 3 FOR j |
≪ 0 1 3 FOR j |
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