bonanova Posted July 20, 2015 Report Share Posted July 20, 2015 What are the next three numbers in this series? 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ... Quote Link to comment Share on other sites More sharing options...
0 hhh3 Posted July 22, 2015 Report Share Posted July 22, 2015 (edited) 138, 150, 180 Edited July 22, 2015 by hhh3 1 Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted July 22, 2015 Report Share Posted July 22, 2015 Whatever you want them to be. There are infinite possibilities. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 22, 2015 Author Report Share Posted July 22, 2015 Hidden ContentOK, what would you want them to be? Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted July 23, 2015 Report Share Posted July 23, 2015 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ... 4+2(n-1)+2(n-1)(n-2)-(4/6)(n-1)(n-2)(n-3)... Continue in this fashion to build all the terms then slap on any scalar along with (n-1)...(n-10) to generate infinite endings Note:n=1 gives 4 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 23, 2015 Author Report Share Posted July 23, 2015 I see what you mean.Kind of like being able to draw an nth-order polynomial through any set of n points. Quote Link to comment Share on other sites More sharing options...
0 dgreening Posted July 23, 2015 Report Share Posted July 23, 2015 I d not see that coming. The classic solution is to find numbers that sandwiched between 2 prime numbers. 1 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 24, 2015 Author Report Share Posted July 24, 2015 OK well since I will define this puzzle not to be an algebra class assignment ...hhh3 has the correct answer. Anyone else see why? Quote Link to comment Share on other sites More sharing options...
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