Division sieve operations in Mathematica

Question

I am just wondering how to easily compute this problem using Mathematica.

My purpose is to extend this to a finite number of factors greater than 4.

Thanks in advance for your help. I know a few Mathematica functions but I don't know what to use in this type of problem.

This seems to be more a question of the underlying math than Mathematica. Could try a search for "inclusion-exclusion methods" to get a start. — Daniel Lichtblau, Commented May 1, 2017 at 22:28
Thanks for the comment Daniel. Indeed to solve the problem, we need to use Exclusion Inclusion Principle. However, computation is tedious for a finitely many facrors. That is why I am asking if there is a mathematica code available in order to get rid of such tedious computations. About your suggestion, I will try to search for that. Thanks again for the comment. I highly appreciate it. — Jr Antalan, Commented May 1, 2017 at 22:36

bill s · Accepted Answer · 2017-05-01 23:03:59Z

2

Here is some Mathematica code to add up the appropriate LCMs:

list = {6, 20, 28, 70};
Total[Flatten[Table[(-1)^(i - 1)/LCM @@@ Subsets[list, {i}], {i, Length[list]}]]]
47/210

If you believe the comments at the linked Mathematics site, this will need to be multiplied by 2.

answered May 1, 2017 at 22:52

bill s

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$\begingroup$ Thanks for your answer bill. I just have a folow up question before trying the code and accept the answer as correct. I for instance I add another factor say 110. What I will do then is to add 110 on the list and change 4 with 5? Am I correct? $\endgroup$
– Jr Antalan
Commented May 1, 2017 at 22:59
$\begingroup$ Yes, just add the new term to the list (I changed the index on the interator). $\endgroup$
– bill s
Commented May 1, 2017 at 23:01
$\begingroup$ Got it bill. Thanks a lot for your answer and help. Till next time. $\endgroup$
– Jr Antalan
Commented May 1, 2017 at 23:03

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ZaMoC · Accepted Answer · 2017-05-01 23:07:55Z

2

list = {6, 20, 28, 70};
S = Drop[Subsets[list], 1];
R = -2*Sum[(-1)^Length[S[[k]]]/LCM @@ S[[k]], {k, 1, Length@S}];
Print[R*100 // N, "%"]

answered May 1, 2017 at 22:58

ZaMoC

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$\begingroup$ Thanks Jenny for your answer. I will try your implement your code. Thanks again. $\endgroup$
– Jr Antalan
Commented May 1, 2017 at 23:01

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2 Answers 2