Guest Posted July 28, 2010 Report Share Posted July 28, 2010 In the following equation, replace x ,y and z with some digits (eg. 1,2,3 etc) and "?" and "!" with some mathematical operator (eg. +-*/), so that the equation hols true xz ? xy = zx ! xy This equation has only two known solutions Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2010 Report Share Posted July 28, 2010 On 7/28/2010 at 12:29 PM, sak_lko said: In the following equation, replace x ,y and z with some digits (eg. 1,2,3 etc) and "?" and "!" with some mathematical operator (eg. +-*/), so that the equation hols true xz ? xy = zx ! xy This equation has only two known solutions Are x, y and z nonzero? If not, then... Reveal hidden contents 01 * 05 = 10 - 05 Quote Link to comment Share on other sites More sharing options...
0 Scudo Posted July 28, 2010 Report Share Posted July 28, 2010 Reveal hidden contents X=0 Y=1 Z=2 !=+ ?=-xz ? xy = zx ! xy Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2010 Report Share Posted July 28, 2010 On 7/28/2010 at 1:41 PM, Scudo said: Reveal hidden contents X=0 Y=1 Z=2 !=+ ?=-xz ? xy = zx ! xy Reveal hidden contents That would be 02 - 01 = 20 + 01 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2010 Report Share Posted July 28, 2010 Reveal hidden contents There is no stipulation that x,y,z must be unique (or that ! and ? are unique). So the entire set of cases where x=y=z and !=? are all valid solutions. Reveal hidden contents 15 + 18 = 51 - 18 27 - 24 = 72 / 24 28 + 27 = 82 - 27 49 - 47 = 94 / 47 If you include leading zeros 01 * 05 = 10 - 05 09 + 06 = 90 / 06 02 + 09 = 20 - 09 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2010 Report Share Posted July 28, 2010 Reveal hidden contents You can replace all with zeros or all with ones or as kevink said make all the same you'd have a true equation Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2010 Report Share Posted July 28, 2010 Reveal hidden contents All ones! Yay one party! (strobe lights and one glyphs dancing) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2010 Report Share Posted July 28, 2010 (edited) If we allow x, y and z and/or the arithmetic operations to not be distinct, and/or allow leading zeroes, we have many solutions. I believe sak_iko meant that x ≠ y, x ≠ z, y ≠ z; x > 0; z > 0; and ? ≠ !. Reveal hidden contents {(1, 8, 5), (+, -)} :: 15 + 18 = 51 - 18 = 33 {(2, 7, 8), (+, -)} :: 28 + 27 = 82 - 27 = 55 {(2, 4, 7), (-, /)} :: 27 - 24 = 72 / 24 = 3 {(4, 7, 9), (-, /)} :: 49 - 47 = 94 / 47 = 2 Edited July 28, 2010 by Dej Mar Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2010 Report Share Posted July 28, 2010 X,Y and Z must be non-zero else there are an infinity of answers and the question is pointless. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 29, 2010 Report Share Posted July 29, 2010 X,Y and Z are non-zero and not equal. I have thought about two solutions Reveal hidden contents 27 - 24 = 72 / 24 49 - 47 = 94 / 47 The other two correct solutions are provided by Kevink and Dejmar Reveal hidden contents 28 + 27 = 82 - 27 15 + 18 = 51 - 18 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 29, 2010 Report Share Posted July 29, 2010 Solved by Kevink and Dej Mar Reveal hidden contents :0 :0 Quote Link to comment Share on other sites More sharing options...
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In the following equation, replace x ,y and z with some digits (eg. 1,2,3 etc) and "?" and "!" with some mathematical operator (eg. +-*/), so that the equation hols true
xz ? xy = zx ! xy
This equation has only two known solutions
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