Guest Posted June 23, 2009 Report Share Posted June 23, 2009 (edited) hi guys, this is my first puzzle. ABCDE*4=EDCBA find "ABCDE". i hope no one posted this puzzle ever. Edited June 23, 2009 by rohit_bd Quote Link to comment Share on other sites More sharing options...
0 Pickett Posted June 23, 2009 Report Share Posted June 23, 2009 I can give an explanation and steps to solving if people want...but it's pretty self-explanatory... 21978 * 4 = 87912 So, ABCDE = 21978 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 23, 2009 Report Share Posted June 23, 2009 A=0 B=0 C=0 D=0 E=0 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 23, 2009 Report Share Posted June 23, 2009 Dempsey Collins , that was really good answer. ur genius! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 23, 2009 Report Share Posted June 23, 2009 pickett ur right. any other possiblities. Quote Link to comment Share on other sites More sharing options...
0 Pickett Posted June 23, 2009 Report Share Posted June 23, 2009 pickett ur right. any other possiblities. If you assume the following about the OP: 1. Each letter is a distinct number (meaning two letters can NOT have the same numerical value) 2. The numbers cannot start with 0 Then, I would say "no" there are no other possibilities. Here's the way to solve this and essentially the proof that there aren't any others: Given: ABCDE*4=EDCBA A must be even and not 0 ABCDE must be less than 25000 So, therefore A must be 2 2BCDE * 4 = EDCB2 BA must be a multiple of 4 B must be 1, 3, 5, 7, or 9 E must be 3 or 8 to result in number ending in 2..but 3 doesn't work...so E must be 8 2BCD8 * 4 = 8DCB2 B must be 0 or 1 so that answer is less than 90000, but can't be 0, since that would result in an answer that is not a multiple of 4...so B = 1 21CD8 * 4 = 8DC12 D must be either 7 or 2 so that 4 * D + 3 = a number ending in 1...(so 4*D must end in 8...aka, D must be 7 or 2) But D can't be 2, since A is 2...so D = 7: 21C78 * 4 = 87C12 Finally, that leaves C being either 0, 3, 4, 5, 6, or 9 well, 4 * C + 3 = number ending in C... so C can not be 0, 3, 4, 5, or 6...which leaves C being 9 Therefore the equation becomes: 21978 * 4 = 87912 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 23, 2009 Report Share Posted June 23, 2009 (edited) Wrote a SQL program and this is the only 5 digit number found. 21978 --Program DECLARE @Flag numeric DECLARE @Flag2 numeric SET @Flag = 10000 WHILE (@Flag < 99999) BEGIN SET @Flag2=@Flag*4 IF(@Flag2=REVERSE(@Flag)) BEGIN PRINT @Flag --BREAK END SET @Flag=@Flag+1 END Edited June 23, 2009 by Terminator Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 23, 2009 Report Share Posted June 23, 2009 Dempsey Collins your answer is wrong. Considering that the problem clearly states that ABCDE*4=EDCBA then the only number is 21978. Why can't it be 00000*4=00000? The condition of existence of this number is that A can't be = to 0. Because that way the number would be reduced to BCDE so 10>A>0 and the only number that meets these conditions is 21978 Oh and don't take it wrong I applaud your efford in trying to solve the puzzle Carry on!!! I was just helping you out Quote Link to comment Share on other sites More sharing options...
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Guest
hi guys, this is my first puzzle.
ABCDE*4=EDCBA
find "ABCDE".
i hope no one posted this puzzle ever.
Edited by rohit_bdLink to comment
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