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Sum multiples of 3 and 5: Difference between revisions
→{{header|MATLAB}} / {{header|Octave}}
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Another alternative is
<lang MATLAB>n=1000; sum(0:3:n-1)+sum(0:5:n-1)-sum(0:15:n-1)</lang>
Of course, its more efficient to use [http://mathforum.org/library/drmath/view/57919.html Gauss' approach] of adding subsequent integers
<lang MATLAB>
n=1e20-1;
n3=floor(n/3);
n5=floor(n/5);
n15=floor(n/15);
(3*n3*(n3+1) + 5*n5*(n5+1) - 15*n15*(n15+1))/2
</lang>
<pre>ans = 2.33333333333333e+39
</pre>
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