dark_magician_92 Posted July 3, 2011 Report Share Posted July 3, 2011 i recently came across this:- 33+43+53=63 i am curious to know if there are more of this kind i.e. sum of cubes of nos. in arithmetic progression = to cube of some other no. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 3, 2011 Report Share Posted July 3, 2011 I put your premise into a spreadsheetI stopped at 32727 and found no occurrence where three consecutive numbers cubed totaled to a perfect cube, which is what I expected.. Quote Link to comment Share on other sites More sharing options...
0 dark_magician_92 Posted July 3, 2011 Author Report Share Posted July 3, 2011 I put your premise into a spreadsheetI stopped at 32727 and found no occurrence where three consecutive numbers cubed totaled to a perfect cube, which is what I expected.. thats ok, but why take consecutive numbers only? i said AP with a common diff. d , not d=1 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 3, 2011 Report Share Posted July 3, 2011 (edited) Okay then ---- 6^3+8^3+10^3=12^3 9^3+12^3+20^3=18^3 12^3+16^3+20^3=24^3 15^3+20^3+25^3=30^3 18^3+24^3+30^3=36^3 That should be enough to find a distinct pattern Sorry for being lazy and not using superscript. Edited July 3, 2011 by thoughtfulfellow Quote Link to comment Share on other sites More sharing options...
0 dark_magician_92 Posted July 3, 2011 Author Report Share Posted July 3, 2011 (edited) Okay then ---- 6^3+8^3+10^3=12^3 9^3+12^3+20^3=18^3 12^3+16^3+20^3=24^3 15^3+20^3+25^3=30^3 18^3+24^3+30^3=36^3 That should be enough to find a distinct pattern Sorry for being lazy and not using superscript. 1st is correct, 2nd shud have 15 cube (too lazy) 3rd ok, 4th ok, 5th ok, yes i observe the pattern thanks a lot Edited July 3, 2011 by dark_magician_92 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 4, 2011 Report Share Posted July 4, 2011 The general equation for 33+43+53 = 63 is [k1/3×3]3+ [k1/3×4]3+[k1/3×5]3 = [k1/3×6]3. Thus, besides 33+ 43+53 = 63 for k=1, there is 63+83+103 = 123 for k=2, 93+123+153 = 183 for k=3, etc. In addition, there is the general equation [k1/3×11]3+[k1/3×12]3+[k1/3×13]3+[k1/3×14]3 = [k1/3×20]3 for 113+123+133+143 = 203 for k=1, 223+243+263+283 = 403 for k=2, 333+363+393+423 = 603 for k=3, etc. Quote Link to comment Share on other sites More sharing options...
0 dark_magician_92 Posted July 4, 2011 Author Report Share Posted July 4, 2011 The general equation for 33+43+53 = 63 is [k1/3×3]3+ [k1/3×4]3+[k1/3×5]3 = [k1/3×6]3. Thus, besides 33+ 43+53 = 63 for k=1, there is 63+83+103 = 123 for k=2, 93+123+153 = 183 for k=3, etc. In addition, there is the general equation [k1/3×11]3+[k1/3×12]3+[k1/3×13]3+[k1/3×14]3 = [k1/3×20]3 for 113+123+133+143 = 203 for k=1, 223+243+263+283 = 403 for k=2, 333+363+393+423 = 603 for k=3, etc. i am giving u a prize Quote Link to comment Share on other sites More sharing options...
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dark_magician_92
i recently came across this:-
33+43+53=63
i am curious to know if there are more of this kind i.e. sum of cubes of nos. in arithmetic progression = to cube of some other no.
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